7 concepts24 worked examples31 exercises5 exam-level6 figures
What are you here for?
13Integration techniques: parts and trigonometric integrals
Two integrals sitting one line apart on a past final: $\int 2x\cos(x^{2})\,dx$ and $\int x\cos x\,dx$. The first is over in ten seconds, because the $2x$ out front is exactly what the substitution $u=x^{2}$ needs. The second looks tamer and there is nothing to substitute: every $u$ you try leaves a stray $x$ standing in the way.
You will be able to look at a product under an integral sign, name in one sentence which move it wants, run that move to a finished answer, and check the answer yourself by differentiating it.
In 60 seconds
Two reversals and one limit: parts undoes the product rule, the undoes a power of sine or cosine, and an integral that runs to infinity is an ordinary integral wrapped in a limit.
$$\int u\,dv=uv-\int v\,du$$
a product of two factors that have nothing to do with each other, one of which gets simpler when differentiated
an endpoint is infinite, or the integrand blows up somewhere on the interval
Three most common mistakes
Losing the minus sign in $uv-\int v\,du$: it sits in front of the whole new integral, and $v$ brings its own sign with it.
in a trigonometric integral and then forgetting to convert what is left with $\sin^{2}x+\cos^{2}x=1$, so the substitution cannot be carried out at all.
Writing $\left[F(x)\right]_{1}^{\infty}=F(\infty)-F(1)$ instead of taking a limit, which quietly treats infinity as a number.
This is week 13 material, so both midterms are behind you: the graded places it can appear are the final, which carries 28 percent of the course, and whatever quizzes are left out of the 10 percent that quizzes carry in total.
How much time do you have?
10 minutes
You leave able to run one pass of parts on a clean product and to peel an odd power in a trigonometric integral. Those two cover most of what an exam asks in this week.
In 60 seconds card, Undoing the product rule, Which factor gets to be u, Peeling the odd power, Formula card
45 minutes
Add the cases that make people lose whole questions: parts applied twice, the integral that comes back to you, both exponents even, and the limit that an infinite endpoint forces.
everything in the 10 minute path, Two passes, and the integral that returns, Definite parts and formulas that feed on themselves, When both powers are even, Integrals with no right-hand end, Full exam-style question, Practice C
full read
The parts that transfer: the reference triangle between the two techniques, the ladder where the scaffolding comes off one rung at a time, and the interleaved set where nothing tells you which technique the question wants.
all blocks in order, Contrast pairs, Scaffolding comes off, Practice A to D, Mistake ledger, Self audit
By the end of this section
Recognise from the shape of an integrand whether a substitution can finish it or whether integration by parts is the only move available.
Choose $u$ and $dv$ so that the traded integral is easier than the one you started with, and say in one sentence why the other choice is worse.
Apply integration by parts more than once, in table form when the polynomial factor dies, and solve for the original integral when it reappears on the right.
Evaluate a definite integral by parts with the handled separately, and derive a reduction formula that lowers an exponent by two.
Integrate $\sin^{m}x\cos^{n}x$ when at least one exponent is odd, by peeling one factor, converting the rest with the Pythagorean identity and substituting.
Integrate $\sin^{m}x\cos^{n}x$ when both exponents are even, using the power reduction identities, and handle a product of sines and cosines of different angles.
Decide whether an improper integral converges by writing it as a limit of ordinary definite integrals, and report a value only when that limit is finite.
Syllabus coverage
7.1
Integration by parts
The formula and what it trades, the choice of $u$, repeated passes and the table that bookkeeps them, the integral that returns, the definite version, and reduction formulas.
covered
7.8
Improper integrals
The two ways an integral becomes improper, the limit definition of each, the family $\int_{1}^{\infty}x^{-p}\,dx$ that decides most questions by comparison, and the interior blow-up that hides in plain sight. The comparison tests themselves come with the next block, where the techniques that produce the missing antiderivatives are also taught.
covered
7.2
Trigonometric integrals
The odd-exponent peel and the even-exponent power reduction, continued in the following concept, with the tangent and secant family and the product-to-sum identities attached there.
Integration by parts is this identity read backwards. Everything in the first half of the block is one rearrangement away from it.
Substitution
$$\displaystyle\int f\bigl(g(x)\bigr)g'(x)\,dx=\int f(u)\,du$$ with $u=g(x)$, $du=g'(x)\,dx$.
It is the technique that has to fail before parts is worth trying, and it is also the last step of almost every trigonometric integral here.
The Pythagorean identity and its two relatives
$\sin^{2}x+\cos^{2}x=1$, and dividing by $\cos^{2}x$ or by $\sin^{2}x$ gives $1+\tan^{2}x=\sec^{2}x$ and $1+\cot^{2}x=\csc^{2}x$.
This is the only tool that converts an even power of one trigonometric function into a polynomial in the other one, which is what makes a substitution possible.
Double angle and power reduction identities
$\sin 2x=2\sin x\cos x$ and $\cos 2x=1-2\sin^{2}x=2\cos^{2}x-1$, which rearrange to $\sin^{2}x=\dfrac{1-\cos 2x}{2}$ and $\cos^{2}x=\dfrac{1+\cos 2x}{2}$.
When both exponents are even there is nothing to peel, and these are the only way in.
The evaluation half of the Fundamental Theorem
If $F'=f$ and $f$ is continuous on $[a,b]$, then $$\displaystyle\int_{a}^{b}f(x)\,dx=F(b)-F(a)$$.
Every definite computation in the block, improper ones included, ends with this line. The improper ones simply take a limit of it.
The antiderivatives we lean on
$$\displaystyle\int e^{ax}\,dx=\frac{e^{ax}}{a}+C$$, $$\displaystyle\int\frac{du}{u}=\ln\lvert u\rvert+C$$, $$\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$$ for $n\neq-1$, $$\displaystyle\int\frac{dx}{1+x^{2}}=\arctan x+C$$.
These are the pieces every technique here hands back to you at the end. The factor $\frac{1}{a}$ in the first one is the single most dropped symbol in the block.
A logarithm loses to any positive power
$$\displaystyle\lim_{t\to\infty}\frac{\ln t}{t^{k}}=0$$ for every $k>0$, which l'Hospital's rule settles in one step.
Improper integrals whose antiderivative contains a logarithm all end on a limit of this shape, and guessing it wrong flips the verdict.
The other four trigonometric functions are sine and cosine in disguise
$\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$ where $\cos x\neq 0$; $\cot x=\dfrac{\cos x}{\sin x}$ and $\csc x=\dfrac{1}{\sin x}$ where $\sin x\neq 0$.
Set up with the derivatives of the trigonometric functions earlier in the term, and needed here twice over. An integrand carrying any of the four has to be rewritten before a route can be chosen at all, since every branch in the trigonometric half of this block is written for $\sin^{m}x\cos^{n}x$. And the same definitions are what tell you that $\ln\lvert\sec x\rvert$ and $-\ln\lvert\cos x\rvert$ are one answer written two ways.
Try it yourself first (3 questions)
1§13.0 — substitution, before we need the thing that replaces it●○○○○
One line of warm-up on the technique you already have. If it is not quick, the substitution rule is the thing to revisit first.
Given
$$\displaystyle\int 2x\,e^{x^{2}}\,dx$$
Find
Evaluate the integral.
Hint 1/4
Look for a factor that is the derivative of something else in the integrand. You are not being asked to be clever.
Hint 2/4
The substitution rule: if the integrand is $f(g(x))\,g'(x)$, then $u=g(x)$ turns the whole thing into $\int f(u)\,du$.
Hint 3/4
Here $g(x)=x^{2}$ and $g'(x)=2x$, which is standing right there. So $u=x^{2}$, $du=2x\,dx$, and $\int 2xe^{x^{2}}\,dx=\int e^{u}\,du$.
Hint 4/4
The answer is $e^{x^{2}}+C$.
Show solution
Substitution rather than anything cleverer, because the derivative of the exponent is already sitting in the integrand as a factor. That is the whole test for whether substitution will work.
Name the inside function and its derivative
$$u=x^{2},\qquad du=2x\,dx$$
the exponent is the only composite piece, and its derivative is the factor already present
$$\int 2xe^{x^{2}}\,dx=\int e^{u}\,du$$
$2x\,dx$ is replaced wholesale by $du$; nothing is left over, which is what makes the substitution complete
$$=e^{u}+C=e^{x^{2}}+C$$
back-substitution, because the question was asked in $x$
Answer $$e^{x^{2}}+C$$
Check
Differentiate: $\frac{d}{dx}e^{x^{2}}=e^{x^{2}}\cdot 2x$, which is the integrand. The chain rule factor coming back out is exactly the $2x$ we consumed.
2§13.0 — the antiderivative that gets a sign wrong more often than any other●○○○○
Half of the sign errors in this whole block come from this one line, so it is worth being sure of before anything else. No substitution is needed; it is a standard form.
Given
$$\displaystyle\int e^{-x}\,dx$$
Find
Which one is the antiderivative?
Hint 1/4
Do not integrate. Differentiate each candidate and see which one gives back $e^{-x}$.
Hint 2/4
For a linear exponent, $\int e^{ax}\,dx=\frac{1}{a}e^{ax}+C$. Read off $a$ before you write anything.
Hint 3/4
Here the integrand is $e^{-x}$, so $a=-1$ and $\frac{1}{a}=-1$.
Hint 4/4
The antiderivative is $-e^{-x}+C$.
Show solution
We use the linear-exponent form rather than a substitution, because the correction factor is a constant either way.
Read the slope of the exponent
$$a=-1$$
the exponent is $-x$, whose derivative is the constant $-1$
dividing by $a$ is what cancels the factor the chain rule will produce
Answer $$-e^{-x}+C$$
Check
A check that needs no integration: $e^{-x}$ is positive, so any antiderivative of it must be increasing, and of the four candidates only $-e^{-x}$ increases.
3§13.0 — a logarithm racing a first power●●○○○
The last block of this section ends on limits of exactly this shape. Both pieces grow without bound, which is why the answer is not immediate.
Given
$$\displaystyle\lim_{t\to\infty}\frac{\ln t}{t}$$
Find
What is the value of the limit?
Hint 1/4
Notice the shape first: both the top and the bottom run off to infinity, so nothing can be read off directly.
Hint 2/4
For an $\frac{\infty}{\infty}$ shape, l'Hospital's rule lets you differentiate the top and the bottom separately and try again.
Hint 3/4
Here that gives $\lim_{t\to\infty}\frac{1/t}{1}=\lim_{t\to\infty}\frac{1}{t}$, and the original ratio was $\frac{\ln t}{t}$.
Hint 4/4
The limit is $0$.
Show solution
L'Hospital is the cheapest route because one differentiation destroys the logarithm entirely; rewriting the ratio by hand would take longer and end in the same place.
Name the shape, then apply the rule
$$\ln t\to\infty,\quad t\to\infty$$
checking the hypothesis before using the rule, since l'Hospital applies only to $\frac{0}{0}$ and $\frac{\infty}{\infty}$
differentiating numerator and denominator separately, which is the rule and not the quotient rule
$$=\lim_{t\to\infty}\frac{1}{t}=0$$
the new limit is no longer indeterminate, so it can be read off
Answer $$0$$
Check
Numbers instead of rules: at $t=10^{6}$ the ratio is $\frac{13.8}{10^{6}}$, about $1.4\times 10^{-5}$. Multiplying $t$ by another million multiplies the numerator by roughly $1.4$ and the denominator by a million.
Any positive power of $t$ beats a logarithm eventually, which is the fact that settles the convergence question at the end of this block.
Notation
symbol
reads as
means
watch out
$\int u\,dv$
the integral of u dee vee
The integrand has been split into a factor called $u$ and everything else, $dv$, which always includes the $dx$.
$dv$ is not a function, it is a function times $dx$. Writing $dv=e^{x}$ instead of $dv=e^{x}\,dx$ loses the $dx$ and with it the ability to antidifferentiate.
$du$, $v$
dee you, vee
$du=u'(x)\,dx$ is what $u$ becomes when differentiated; $v$ is any antiderivative of $dv$.
One of the two is differentiated and the other is antidifferentiated. Doing the same operation to both is the most common way the method produces nonsense.
$\bigl[uv\bigr]_{a}^{b}$
u vee evaluated from a to b
$u(b)v(b)-u(a)v(a)$, the boundary term of a definite integration by parts.
It is a number, not a function, and it is not integrated again.
$I_{n}$
I sub n
A whole family of integrals indexed by an exponent, so that one formula can relate the member at $n$ to the member at $n-2$.
$I_{n}$ is not a number until $n$ is a number. A reduction formula is an equation between two members of the family.
$\sin^{m}x$
sine to the m of x
$(\sin x)^{m}$, the sine taken first and then raised to the power.
It is not $\sin(x^{m})$, and for $m=-1$ the notation would mean the reciprocal here, not the inverse function; we write $\arcsin$ for the inverse to keep the two apart.
$$\displaystyle\int_{a}^{\infty}f(x)\,dx$$
the integral of f from a to infinity
Shorthand for $\lim_{t\to\infty}\int_{a}^{t}f(x)\,dx$, and it means nothing else.
The symbol $\infty$ inside it is not a point you can substitute into an antiderivative.
$converges, diverges$
converges, diverges
The defining limit exists and is finite, or it does not.
"Diverges" covers both the case where the limit runs off to infinity and the case where it oscillates and settles nowhere.
$+C$
plus C
The arbitrary constant on an indefinite integral.
It belongs on the final line of an indefinite answer and is left off intermediate lines inside a table, where it would be absorbed anyway.
Conventions used here
The constant of integration
Every indefinite integral here ends in $+C$, and intermediate lines inside a table or a reduction step are written without it. When a step produces a fixed number, for instance the $-1$ that falls out of a boundary evaluation, we absorb it into $C$ and say so on the line where it happens.
Dropping a constant and absorbing a constant look identical on paper, and only one of them is legitimate.
Absolute value bars in logarithms
$\int\frac{du}{u}=\ln\lvert u\rvert+C$ carries bars. We keep them unless the argument is positive for every value in play, as it is in $\ln(1+x^{2})$, where writing bars would only suggest that we do not know why they are usually there.
A missing bar is invisible until the interval crosses zero, and then the answer is undefined rather than merely wrong.
Every angle is in radians
Every trigonometric function in this block takes a radian argument, and every limit of integration written with $\pi$ in it is a radian measure.
The antiderivative formulas used throughout are false in degrees, and nothing in the written work would show it.
Reporting an integral that runs to infinity
The verdict is either the sentence "converges to $L$" with the number $L$ written down, or the sentence "diverges". We write "diverges to $\infty$" when the limit runs off in one direction, because that carries more information than "the limit does not exist", but $\infty$ is never handled as if it were a number.
Treating $\infty$ as a number is exactly how a integral acquires a finite value on an exam paper.
Which antiderivative of dv we take
When we antidifferentiate $dv$ to get $v$ we take the simplest one, with no constant attached. Any other choice cancels out of the final answer, so the simplest choice is free.
Carrying a $+K$ through $uv-\int v\,du$ produces two extra terms that annihilate each other, which costs a line and gains nothing.
Answers stay exact
Answers are left as exact expressions, so $\frac{\pi}{4}-\frac{1}{2}\ln 2$ rather than $0.439$. A decimal appears only when the point of the line is a size comparison.
An exact form can be checked by differentiating it; a decimal cannot.
13.1Undoing the product rule
Trades a product integral for an easier one, $uv$ minus a new integral; use it the moment substitution fails.
The hook left us holding $\int x\cos x\,dx$ with nothing to substitute. The way out is to read a differentiation rule backwards.
Solvable with what we have
$\int 2x\cos(x^{2})\,dx$ — the $2x$ is exactly $du$ for $u=x^{2}$
$\int xe^{x^{2}}\,dx$ — the same trick with half a $du$
$\int\frac{\ln x}{x}\,dx$ — the $\frac{1}{x}$ is $du$ for $u=\ln x$
$\int\cos^{4}x\sin x\,dx$ — the lone $\sin x$ is $-du$ for $u=\cos x$
Not solvable yet
$\int x\cos x\,dx$
$\int xe^{x}\,dx$
$\int\ln x\,dx$
$\int x\arctan x\,dx$
Antidifferentiate the two factors separately and multiply: $\int x\cos x\,dx\stackrel{?}{=}\frac{x^{2}}{2}\sin x$. Differentiate the guess to test it: $\frac{d}{dx}\left[\frac{x^{2}}{2}\sin x\right]=x\sin x+\frac{x^{2}}{2}\cos x$. The integrand was $x\cos x$. Not close, and not fixable by a constant.
Why it fails
Differentiating a product hands back two terms, so undoing it has to account for two terms. Treating each factor on its own accounts for one of those terms and throws the other away. The repair is not a better guess. It is to integrate the product rule itself and see what falls out.
TheoremIntegration by parts
Conditions
$u$ and $v$ are differentiable on the interval in question, with continuous derivatives there
$dv$ carries the $dx$: everything that is not $u$, the differential included, is $dv$
$v$ is one antiderivative of $dv$, and we always take the simplest one, with no constant attached
Differentiate one factor, antidifferentiate the other, write down the product of what you have, and subtract the integral of the new pair. Notice what this is not: it does not evaluate anything. It exchanges the integral you have for a different integral, and the entire skill is in making that exchange a trade up.
Where the formula comes from
Start from the product rule, $\frac{d}{dx}\left[u v\right]=u'v+uv'$.
Integrate both sides with respect to $x$. The left side integrates to $uv$, since $uv$ is by definition an antiderivative of its own derivative, so $uv=\int u'v\,dx+\int uv'\,dx$.
Move one term across: $\int uv'\,dx=uv-\int u'v\,dx$.
Writing $dv=v'\,dx$ and $du=u'\,dx$ turns that into the boxed line. Nothing has been assumed beyond the product rule, which is why the formula is available for any pair of differentiable factors, useful or not.
The identity drawn once, for $u=\ln x$ between $1$ and $e$. The whole rectangle is the boundary term $\textcolor{#6f42c1}{\left[x\ln x\right]_{1}^{e}=e}$. The curve cuts it into the integral we want, $\textcolor{#1f6feb}{\int_{1}^{e}\ln x\,dx=1}$, and the integral we trade it for, $\textcolor{#d1690a}{e-1}$. Parts is the statement that one of those two areas is the rectangle minus the other.
Looks like this, but is not
$\int x\cos(x^{2})\,dx$ is also an algebraic factor multiplied by a trigonometric one, so it looks like the same kind of job.
It is not. The $x$ in front is half the derivative of the $x^{2}$ inside, so $u=x^{2}$ finishes it in a line and gives $\frac{1}{2}\sin(x^{2})+C$. What makes an integral a parts problem is not the presence of a product. It is that the two factors are strangers to each other: neither one is the derivative of anything sitting inside the other.
The integral the hook got stuck on: ∫ x cos x dx
Nothing to substitute, two factors that have nothing to do with each other. This is the shape parts was built for.
Given
$$\displaystyle\int x\cos x\,dx$$
Find
the antiderivative
Solution
We let $x$ be $u$ rather than $\cos x$, because differentiating $x$ destroys it while differentiating $\cos x$ only changes its name. The factor that gets simpler under differentiation is the one worth differentiating.
Write all four pieces before integrating anything
$$u=x,\qquad dv=\cos x\,dx$$
the algebraic factor is the one that improves when differentiated; the trigonometric one is no harder to antidifferentiate than to differentiate, so it loses nothing by being $dv$
$$du=dx,\qquad v=\sin x$$
one differentiation and one antidifferentiation, never the same operation twice
Make the trade and look at what you got
$$\int x\cos x\,dx=x\sin x-\int\sin x\,dx$$
the formula, with the minus sign covering the whole new integral
$$=x\sin x-(-\cos x)+C=x\sin x+\cos x+C$$
$\int\sin x\,dx=-\cos x$, and the outer minus turns it into a plus; this is where the sign is most often lost
Answer $$x\sin x+\cos x+C$$
Check
Differentiate the answer: $\frac{d}{dx}\left[x\sin x+\cos x\right]=\sin x+x\cos x-\sin x=x\cos x$. The two $\sin x$ terms cancelling is the signature of a correct parts answer.
One pass. Differentiating $x$ once removes it, so there is nothing left to iterate on.
The trade was worth making for one reason only: $\int\sin x\,dx$ is easier than $\int x\cos x\,dx$, and that comparison is what you check after every pass.
A single factor split in two: ∫ ln x dx
There is only one factor here, so at first glance there is nothing to split. That impression is what the example is for.
Given
$$\displaystyle\int\ln x\,dx$$, on $x>0$
Find
the antiderivative
Solution
Parts rather than substitution, because there is no inner function whose derivative is present. The move is to write the integrand as $\ln x$ times $1$, which costs nothing and creates the second factor the method needs.
Manufacture the second factor
$$\int\ln x\,dx=\int(\ln x)(1)\,dx$$
multiplying by $1$ changes nothing, and it is the only way to have a $dv$ at all
$$u=\ln x,\qquad dv=dx$$
$\ln x$ has no elementary antiderivative we know yet, so it cannot be $dv$; that settles the choice without any heuristic
$$du=\frac{dx}{x},\qquad v=x$$
the derivative of the logarithm is algebraic, which is the simplification we were buying
the formula; the $x$ from $v$ and the $\frac{1}{x}$ from $du$ are about to cancel, which is the whole payoff
$$=x\ln x-\int 1\,dx=x\ln x-x+C$$
the traded integral is now the easiest one there is
Answer $$x\ln x-x+C$$
Check
Differentiate: $\ln x+x\cdot\frac{1}{x}-1=\ln x+1-1=\ln x$. As a second, independent check, the figure above says $\int_{1}^{e}\ln x\,dx$ should be $1$, and this antiderivative gives $(e-e)-(0-1)=1$.
Any integrand can be written as itself times $1$, and that is the standard move whenever the derivative of the integrand is simpler than the integrand.
Checkpoint
§13.1 — telling a substitution from a parts problem●●○○○
Thirty seconds, no work on paper. Three of these four are finished by a substitution; one of them is not, and that one is the reason this section exists.
Given
$\int x\,\cos x\,dx$
$\int 2x\cos(x^{2})\,dx$
$\int\cos^{3}x\sin x\,dx$
$\int\cos x\sin^{4}x\,dx$
Find
Which one has no substitution that finishes it?
Hint 1/4
Do not integrate anything. For each one, ask a single question: is one factor the derivative of something inside the other factor?
Hint 2/4
A substitution $u=g(x)$ works when the integrand is $f(g(x))$ multiplied by $g'(x)$, up to a constant.
Hint 3/4
In $\int 2x\cos(x^{2})\,dx$ take $g=x^{2}$; in $\int\cos^{3}x\sin x\,dx$ take $g=\cos x$; in $\int\cos x\sin^{4}x\,dx$ take $g=\sin x$. Now try to name a $g$ for $\int x\cos x\,dx$.
Hint 4/4
The one with no substitution is $\int x\cos x\,dx$.
Show solution
We test for substitution before reaching for parts, because a substitution is always cheaper when it exists and finding out costs one line per integrand.
Test each candidate
$$g=x^{2}\ \Rightarrow\ g'=2x$$
present as a factor in the second integrand, so a substitution finishes it
$$g=\cos x\ \Rightarrow\ g'=-\sin x$$
present up to sign in the third, and a constant multiple is allowed
$$g=\sin x\ \Rightarrow\ g'=\cos x$$
present in the fourth
$$\int x\cos x\,dx$$
the only inside function is $x$ itself, whose derivative is $1$; the leftover $x$ is not a constant, so no substitution clears the integrand
Answer $$\int x\cos x\,dx$$
Check
A different way to see it: substitution never lowers the degree of an algebraic factor sitting outside the composition, and in $\int x\cos x\,dx$ only differentiation can remove the $x$.
⚠ Antidifferentiating the two factors separately
the sum rule really does let you integrate term by term, and the habit gets carried across to products, where it is false
wrong$$\int x\cos x\,dx=\frac{x^{2}}{2}\sin x+C$$
right$$\int x\cos x\,dx=x\sin x+\cos x+C$$
⚠ Letting the minus sign cover only the first term
the formula is usually written on one line, and the minus looks like it belongs to $v$ rather than to the whole integral that follows
wrong$$\int x\cos x\,dx=x\sin x+\int\sin x\,dx$$
right$$\int x\cos x\,dx=x\sin x-\int\sin x\,dx$$
13.2Which factor gets to be u
LIATE ranks the five factor types so you pick the $u$ that simplifies when differentiated, before writing anything down.
The formula is legal whichever way you split the integrand. Only one of the two splits is useful, and the gap between them is a page of work.
MethodLIATE, a priority order for choosing u
Conditions
it is a habit with a good reason behind it, not a theorem, and integrands can be built to defeat it
when two of the five classes appear, the one earlier in the list becomes $u$ and everything else becomes $dv$
you must be able to antidifferentiate whatever you called $dv$, so a $dv$ you cannot handle overrides the order
Find which of the five classes the factors belong to and let the earliest one on the list be $u$. The order is not arbitrary: it ranks the classes by how much they give up when you differentiate them. A logarithm collapses into an algebraic fraction, an inverse trigonometric function does the same, a power drops a degree, while a sine and an exponential simply turn into other functions of the same size. You differentiate the one with something to lose.
Why the order is what it is
The traded integral is $\int v\,du$, so you win exactly when $du$ is simpler than $u$ and $v$ is no worse than $dv$.
Differentiating $\ln x$ gives $\frac{1}{x}$, an algebraic function: a whole class simpler. Differentiating $\arctan x$ gives $\frac{1}{1+x^{2}}$, again algebraic. Differentiating $x^{n}$ gives $nx^{n-1}$, one degree down, so repeating it enough times ends at a constant. Differentiating $\sin x$ gives $\cos x$ and differentiating $e^{x}$ gives $e^{x}$: no progress at all, ever.
So the list is just those four observations sorted by size of gain. The last two classes are at the end not because they are difficult but because they are indifferent, which makes them the right things to antidifferentiate.
Looks like this, but is not
$\int x^{5}e^{x^{3}}\,dx$ has an algebraic factor and an exponential one, so LIATE says $u=x^{5}$, $dv=e^{x^{3}}\,dx$.
That choice is unusable, because $e^{x^{3}}$ has no elementary antiderivative, so there is no $v$ to write down. The integrand is not really a parts problem in the first place: substituting $t=x^{3}$ turns $x^{5}\,dx$ into $\frac{1}{3}t\,dt$ and leaves $\frac{1}{3}\int te^{t}\,dt$, which then is one. LIATE ranks the choices; it does not check whether $dv$ can be antidifferentiated, and that check comes first.
choice
$du$ and $v$
the traded integral $\int v\,du$
verdict
$u=\ln x$, $dv=x\,dx$
$du=\frac{dx}{x}$, $v=\frac{x^{2}}{2}$
$\frac{1}{2}\int x\,dx$
a bare power, done in one line
$u=x$, $dv=\ln x\,dx$
$du=dx$, $v=x\ln x-x$
$\int(x\ln x-x)\,dx$
contains $\int x\ln x\,dx$ again, the integral we started from
The second row is not merely slower. Its traded integral contains the original integral, so the trade goes backwards: you would have to integrate by parts once just to find $v$, and then find the same problem waiting on the right.
Both classes present: ∫ x ln x dx
An algebraic factor and a logarithmic one, so the list has something to say.
Given
$$\displaystyle\int x\ln x\,dx$$, on $x>0$
Find
the antiderivative
Solution
L beats A, so $u=\ln x$. The reverse choice needs $\int\ln x\,dx$ before it can even start, which is a separate integration by parts.
Split by the list
$$u=\ln x,\qquad dv=x\,dx$$
logarithmic beats algebraic, and $x\,dx$ is trivial to antidifferentiate
$$du=\frac{dx}{x},\qquad v=\frac{x^{2}}{2}$$
the simplest antiderivative, with no constant, by the convention we set at the top
the logarithm is gone from the integral, which was the point of choosing it as $u$
Answer $$\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C$$
Check
Differentiate: $x\ln x+\frac{x^{2}}{2}\cdot\frac{1}{x}-\frac{x}{2}=x\ln x+\frac{x}{2}-\frac{x}{2}=x\ln x$. A second check with numbers: on $[1,2]$ the answer gives $2\ln 2-1-(0-\frac{1}{4})=2\ln 2-\frac{3}{4}\approx 0.636$, and the integrand runs from $0$ to $2\ln 2\approx 1.39$ over an interval of length $1$, so a value around $0.6$ is the right size.
One pass, one cancellation.
An inverse trigonometric factor: ∫ x arctan x dx
Now the list picks the inverse trigonometric factor, and the traded integral needs one algebraic step of its own.
Given
$$\displaystyle\int x\arctan x\,dx$$
Find
the antiderivative
Solution
I beats A, so $u=\arctan x$. Reversing the roles would need $\int\arctan x\,dx$ first, which is itself a parts problem, so the reversed choice is strictly more expensive.
the numerator has the same degree as the denominator, so it must be split before any standard form applies; adding and subtracting $1$ is cheaper than long division here
Differentiate the compact form: $x\arctan x+\frac{x^{2}+1}{2}\cdot\frac{1}{1+x^{2}}-\frac{1}{2}=x\arctan x+\frac{1}{2}-\frac{1}{2}=x\arctan x$. The two constant halves cancelling is the sign that the algebraic detour was done correctly.
Whenever $\frac{1}{1+x^{2}}$ meets a power of $x$, rewrite the numerator in terms of the denominator, as in $x^{2}=(1+x^{2})-1$.
Checkpoint
§13.2 — splitting an integrand the useful way●●○○○
You are about to integrate a product of a power and a logarithm. Only the split is being asked for, not the answer.
Given
$$\displaystyle\int x^{2}\ln x\,dx$$
Find
Which split should you make?
Hint 1/4
You are choosing which factor to differentiate. Ask which of the two gets simpler when you do.
Hint 2/4
The order is logarithmic, then inverse trigonometric, then algebraic, then trigonometric, then exponential; the earliest class present becomes $u$.
Hint 3/4
Here the classes present are logarithmic, $\ln x$, and algebraic, $x^{2}$. The list puts logarithmic first, so $u=\ln x$ and everything else, $x^{2}\,dx$, is $dv$.
Hint 4/4
The split is $u=\ln x$ with $dv=x^{2}\,dx$.
Show solution
We check what $dv$ demands before we check the list, because a $dv$ with no known antiderivative kills a choice outright, and here that alone eliminates the second option.
Differentiate the result: $x^{2}\ln x+\frac{x^{3}}{3}\cdot\frac{1}{x}-\frac{x^{2}}{3}=x^{2}\ln x$. The two $\frac{x^{2}}{3}$ terms cancel, as they must.
⚠ Picking a dv you cannot antidifferentiate
the list is memorised as a rule and applied before the more basic question of whether $v$ exists at all
When one pass is not enough: tabulate until the polynomial dies, or solve algebraically for the integral that reappears.
One pass finished the last three examples because differentiating $u$ destroyed it in a single step. When $u$ is $x^{3}$ that takes three steps, and when $u$ is $\sin x$ it never happens at all.
RuleRepeating the trade
Conditions
each pass keeps the same $u$-and-$dv$ orientation as the first one; swapping halfway undoes the previous pass exactly
the table form needs a $u$ whose derivatives reach $0$, which means a polynomial
the returning form needs the reappearing integral to come back with a coefficient different from $1$, otherwise the equation says $0=0$
In the table form, $v_{1},v_{2},v_{3},\dots$ are the repeated antiderivatives of $dv$ and the signs alternate starting with a plus; you stop when the left column reaches zero. In the returning form, two passes bring the original integral back on the right-hand side, and at that point you stop integrating and start solving: treat the integral as an unknown and move it across.
Why the table is just parts written sideways
Apply parts once to $\int u\,dv$: you get $uv_{1}-\int v_{1}\,du$.
Apply it again to $\int v_{1}u'\,dx$, this time with $u'$ as the new $u$: that gives $u'v_{2}-\int v_{2}u''\,dx$. Substituting it back and keeping track of the outer minus produces $uv_{1}-u'v_{2}+\int v_{2}u''\,dx$.
Each further pass appends one more term and flips the sign of the remaining integral, which is exactly the alternating pattern. When $u$ is a polynomial of degree $n$, the derivative $u^{(n+1)}$ is zero, the last integral is $\int 0\,dx$, and the process stops with no remainder. The table is that computation with the bookkeeping laid out in two columns instead of being written in one long line.
Three passes of parts on $\int x^{3}e^{-x}\,dx$, written as two columns. The left column, $\textcolor{#1f6feb}{u\text{ and its derivatives}}$, is walked down until it hits zero; the right column, $\textcolor{#d1690a}{dv\text{ and its antiderivatives}}$, is walked down the same number of rows. Each diagonal arrow is one product, and the signs alternate starting with a plus.
Looks like this, but is not
$\int x^{2}\ln x\,dx$ has a polynomial factor, so it looks like a table problem: differentiate $x^{2}$ down to zero and antidifferentiate $\ln x$ three times.
The table needs the polynomial to be $u$, and here the list says $u=\ln x$ instead, precisely because $\ln x$ is the factor with something to lose. The table is not a separate technique; it is a layout for the case where the polynomial is the one being differentiated. When the polynomial belongs on the other side, there is nothing to lay out, and one ordinary pass finishes the job anyway.
sign
$u$ and its derivatives
$dv$ and its antiderivatives
the diagonal product
$+$
$x^{3}$
$e^{-x}$
$-x^{3}e^{-x}$
$-$
$3x^{2}$
$-e^{-x}$
$-3x^{2}e^{-x}$
$+$
$6x$
$e^{-x}$
$-6xe^{-x}$
$-$
$6$
$-e^{-x}$
$-6e^{-x}$
$0$
$e^{-x}$
stop
Read the products off the diagonals: row one on the left against row two on the right, and so on. Every one of the four came out negative here, which happens because the sign of the antiderivative alternates in step with the sign in the first column. Adding them gives $-e^{-x}\left(x^{3}+3x^{2}+6x+6\right)$.
A cubic against an exponential: ∫ x³ e⁻ˣ dx
Three passes of parts, written once in the table above and now read off it.
Given
$$\displaystyle\int x^{3}e^{-x}\,dx$$
Find
the antiderivative
Solution
The table rather than three separate passes, because the only thing three passes would add is three chances to lose a sign. The choice of $u$ is forced: differentiating $x^{3}$ terminates, differentiating $e^{-x}$ never does.
Walk both columns down
$$u:\ x^{3}\to 3x^{2}\to 6x\to 6\to 0$$
four differentiations kill a cubic, and the zero row is the stopping signal
Differentiate the factored form: $e^{-x}\left(x^{3}+3x^{2}+6x+6\right)-e^{-x}\left(3x^{2}+6x+6\right)=x^{3}e^{-x}$. Everything except the leading cube cancels, which is the pattern to expect whenever a polynomial has been walked down to zero.
Three passes, compressed into one table with four diagonal products.
The coefficients in the bracket are the left column read downwards, which is a free check on any table answer.
The integral that returns: ∫ e⁻ˣ sin x dx
Neither factor gets simpler when differentiated, so nothing terminates. Two passes bring the original integral back, and that turns out to be the point rather than a failure.
Given
$$\displaystyle I=\int e^{-x}\sin x\,dx$$
Find
the antiderivative
Solution
We keep $u$ trigonometric on both passes. Swapping the roles on the second pass would retrace the first pass backwards and land on the true but useless statement $I=I$, which is the single most common way this example is spoiled.
Differentiate: $\frac{e^{-x}}{2}(\sin x+\cos x)-\frac{e^{-x}}{2}(\cos x-\sin x)=e^{-x}\sin x$. The cosine terms cancel and the sine terms add, which is the check worth doing because a wrong sign inside the bracket survives everything else.
Two passes and one line of algebra. No third pass exists; a third pass returns you to the start.
Recognising the return as the finish line rather than a dead end is the transferable part, and it comes back at $\int\sec^{3}x\,dx$.
Checkpoint
§13.3 — what to do when the integral reappears●●●○○
A student integrates $\int e^{2x}\cos 3x\,dx$ by parts twice, keeping the same orientation both times, and arrives at $I=A-\frac{9}{4}I$ where $A$ is a combination of exponentials and trigonometric functions with no integral in it.
Given
$$I=\displaystyle\int e^{2x}\cos 3x\,dx$$
after two passes: $I=A-\frac{9}{4}I$, with $A$ containing no integral
Find
What is the next move?
Hint 1/4
Stop thinking of $I$ as a task to be finished and look at the line as an ordinary equation with one unknown in it.
Hint 2/4
If $I=A-kI$ with $k\neq-1$, then $(1+k)I=A$ and $I=\frac{A}{1+k}$.
Hint 3/4
Here $k=\frac{9}{4}$, so $1+k=\frac{13}{4}$ and the line reads $\frac{13}{4}I=A$.
Hint 4/4
Solve for $I$: it equals $\frac{4A}{13}$, plus the constant.
Show solution
Algebra rather than another pass, because a third pass reproduces the same cycle and the equation already contains enough information to finish.
Collect the unknown
$$I+\frac{9}{4}I=A$$
moving every occurrence of the unknown to one side, as with any linear equation
$$\frac{13}{4}I=A$$
the coefficient is not $1$, which is exactly the condition that makes the equation informative
$$I=\frac{4A}{13}+C$$
the arbitrary constant returns at the end, having been suppressed while $I$ was a single symbol
Answer $$I=\frac{4A}{13}+C$$
Check
Check against the closed form. Two passes give $A=\frac{e^{2x}}{2}\cos 3x+\frac{3}{4}e^{2x}\sin 3x$, and multiplying that by $\frac{4}{13}$ gives $\frac{e^{2x}\left(2\cos 3x+3\sin 3x\right)}{13}$, which is the standard result $\frac{e^{ax}\left(a\cos bx+b\sin bx\right)}{a^{2}+b^{2}}$ with $a=2$ and $b=3$.
⚠ Swapping the orientation on the second pass
the second pass looks like a fresh problem, and the list is consulted again as if the first pass had not happened
13.4Definite parts, and formulas that feed on themselves
Parts with limits evaluates $uv$ at the endpoints immediately, and one pass yields reduction formulas that drop an exponent by two.
Every example so far was indefinite. With limits attached, one of the two pieces stops being a function and becomes a number, and that changes what you are allowed to do with it.
RuleParts with limits, and reduction formulas
Conditions
$u$ and $v$ have continuous derivatives on the closed interval $[a,b]$
the boundary term is evaluated at both endpoints and subtracted, and it is a number from that moment on
a reduction formula is produced by one pass of parts in which $du$ lowers the exponent, and it is only useful if the recursion reaches a base case you can integrate
When the interval is improper the boundary term is not a number: $\bigl[uv\bigr]_{a}^{t}$ stays a function of the endpoint letter, and taking its limit is a separate step that is usually indeterminate at first sight and is settled by l'Hospital's rule or by a growth limit proved earlier in the same question. The integral can converge perfectly well while that term looks indeterminate.
The first line says: do exactly what you did before, but evaluate the product part at the two endpoints straight away and keep the limits on the integral that is left. The second line is what one pass of parts produces when the integrand is a high power of sine, and it says that the exponent can always be pushed down by two at the cost of one explicit term. Repeat it and the exponent walks down to $0$ or $1$, where the integral is known.
Where the reduction formula comes from
Write $\sin^{n}x=\sin^{n-1}x\cdot\sin x$ and take $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so that $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$.
One pass gives $I_{n}=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\cos^{2}x\,dx$.
Now use $\cos^{2}x=1-\sin^{2}x$ inside that integral. It splits into $(n-1)I_{n-2}-(n-1)I_{n}$, and the original integral has reappeared, exactly as in the returning case.
Collecting gives $nI_{n}=-\sin^{n-1}x\cos x+(n-1)I_{n-2}$, and dividing by $n$ is the boxed line. The identity $\cos^{2}=1-\sin^{2}$ is doing the real work here; parts only creates the opening for it.
What the reduction formula does over $\left[0,\frac{\pi}{2}\right]$, where $\textcolor{#1f6feb}{J_{n}=\int_{0}^{\pi/2}\sin^{n}x\,dx}$. The boundary term $-\sin^{n-1}x\cos x$ is zero at both endpoints, so each step is nothing but a multiplication by a fraction, all the way down to the $\textcolor{#6f42c1}{\text{base case }J_{0}=\frac{\pi}{2}}$.
Looks like this, but is not
In $\int_{0}^{1}xe^{-x}\,dx$ it looks safe to evaluate the boundary term at the limits, get a number, and then carry on integrating the remaining integral without limits.
It is not safe, because the line then contains a number added to a function of $x$, and the final answer will be a function where a number was asked for. Either keep the limits on everything, or drop the limits from everything, work out the whole antiderivative, and evaluate at the very end. Mixing the two is the most common way a correct method produces an answer with an $x$ in it.
A definite integral with an inverse trigonometric factor: ∫₀¹ arctan x dx
One factor, limits attached, and a boundary term that is a genuine number rather than a symbol.
Given
$$\displaystyle\int_{0}^{1}\arctan x\,dx$$
Find
the exact value
Solution
We take $u=\arctan x$ and $dv=dx$, the same manufactured split as with $\ln x$, because the derivative of $\arctan$ is algebraic while $\arctan$ itself is not something we can antidifferentiate yet.
boundary term minus the traded integral, both now numbers
Answer $$\frac{\pi}{4}-\frac{1}{2}\ln 2$$
Check
A size check rather than a repeat of the algebra: $\arctan$ climbs from $0$ to $\frac{\pi}{4}\approx 0.785$ over an interval of length $1$, so the value must lie between those two numbers, and $0.785-0.347=0.439$ does.
Evaluate the boundary term the moment you write it, and check that a definite answer is a number with no $x$ left in it.
Using the reduction formula: ∫₀^{π/2} sin⁴x dx
The formula was derived above; here it is used, and the endpoints make it unusually clean.
The reduction formula rather than the double-angle identity twice: both work, but the recursion is three short lines and leaves no room for an algebra slip in the expansion of $\left(\frac{1-\cos 2x}{2}\right)^{2}$.
at $x=\frac{\pi}{2}$ the factor $\cos x$ is zero, and at $x=0$ the factor $\sin^{n-1}x$ is zero for every $n\ge 2$; this is the only reason the recursion is this clean
$$J_{n}=\frac{n-1}{n}J_{n-2}$$
with the explicit term gone, one step of the recursion is a single multiplication
An independent estimate: $\sin^{4}x\le\sin^{2}x$ on the interval, and $J_{2}=\frac{\pi}{4}\approx 0.785$, so $J_{4}$ has to be smaller than that. The answer is $\frac{3\pi}{16}\approx 0.589$, smaller as required, and larger than $J_{6}$, which the same recursion gives as $\frac{5}{6}J_{4}\approx 0.491$.
Two multiplications, once the boundary term has been shown to vanish.
A reduction formula converts an integration problem into an arithmetic one, at the price of deriving the formula once.
Checkpoint
§13.4 — evaluating the boundary term●●○○○
You are computing $\int_{0}^{1}xe^{-x}\,dx$ by parts with $u=x$ and $dv=e^{-x}\,dx$. Only the boundary term is being asked for, not the whole integral.
Given
$$\displaystyle\int_{0}^{1}xe^{-x}\,dx$$
$u=x$, $dv=e^{-x}\,dx$
Find
What is the value of $\bigl[uv\bigr]_{0}^{1}$?
Hint 1/4
The boundary term is a plain evaluation of a product at two numbers. No integration is involved in this question at all.
Hint 2/4
$\bigl[uv\bigr]_{0}^{1}=u(1)v(1)-u(0)v(0)$, where $v$ is the antiderivative of $dv$.
Hint 3/4
Here $u=x$ and $v=-e^{-x}$, so $uv=-xe^{-x}$, and you are evaluating that at $x=1$ and at $x=0$.
Hint 4/4
The boundary term is $-e^{-1}$.
Show solution
We write $v$ down on its own line before multiplying, because the minus sign that $e^{-x}$ picks up on antidifferentiation is the whole content of this question.
Find v, then evaluate
$$v=\int e^{-x}\,dx=-e^{-x}$$
the linear exponent contributes a factor $\frac{1}{-1}$
$$uv=-xe^{-x}$$
the product, kept as one expression so that both endpoints go into the same formula
the factor $x$ makes the lower endpoint vanish, which is why only one term survives
Answer $$-e^{-1}$$
Check
A sanity check on the sign: the full integral $\int_{0}^{1}xe^{-x}\,dx$ is positive, and it equals $-e^{-1}+\int_{0}^{1}e^{-x}\,dx=-e^{-1}+1-e^{-1}=1-\frac{2}{e}\approx 0.264$. A positive total built from a negative boundary term is exactly what a decreasing $v$ produces.
⚠ Leaving the limits off the boundary term
the indefinite habit is strong, and $uv$ looks like part of the answer rather than a number to be evaluated at once
When either exponent is odd, peel one factor off as $du$, convert the rest with $\sin^2x+\cos^2x=1$, and substitute.
Parts trades one integral for another. The next family does something different: it manufactures a substitution where none existed, and the manufacturing tool is the identity $\sin^{2}x+\cos^{2}x=1$.
Solvable with what we have
$\int\cos^{4}x\sin x\,dx$ — one lone $\sin x$, so $u=\cos x$ and $du=-\sin x\,dx$
$\int\sin^{3}x\cos x\,dx$ — one lone $\cos x$, so $u=\sin x$
$\int\tan x\sec^{2}x\,dx$ — the same idea with $u=\tan x$
Not solvable yet
$\int\sin^{3}x\cos^{2}x\,dx$
$\int\cos^{5}x\,dx$
$\int\sin^{2}x\cos^{2}x\,dx$
Push $u=\cos x$ through $\int\sin^{3}x\cos^{2}x\,dx$ anyway. Then $du=-\sin x\,dx$, which uses up one of the three sine factors, and the integral becomes $-\int\sin^{2}x\cdot u^{2}\,du$. There is still a $\sin^{2}x$ inside, and $\sin^{2}x$ is not written in terms of $u$, so the substitution is not finished and cannot be integrated.
Why it fails
A substitution is complete only when every trace of the old variable is gone. What survived here is an even power of sine, and that turns out to be the good case rather than the fatal one: $\sin^{2}x=1-\cos^{2}x=1-u^{2}$, so it can be converted after all. The failure is not in the substitution; it is in stopping one line too early.
MethodPeel one factor, convert the rest
Conditions
at least one of the two exponents is odd; if both are even this method does not apply at all
you peel a factor of the function whose exponent is odd, and substitute for the other function
what is left after peeling has an even exponent, which is what makes the Pythagorean conversion possible
a negative odd exponent still counts as odd, but the conversion then returns a rational function of $u$ rather than a polynomial: one factor in the denominator divides out term by term, anything heavier waits for partial fractions
If the sine exponent is odd, set one $\sin x$ aside to serve as the differential, rewrite the even number of sines that remain as a polynomial in cosine, and substitute $u=\cos x$. If the cosine exponent is odd, do the mirror image. Either way the integral turns into a polynomial in $u$ as long as both exponents are non-negative, and a polynomial is something you have been able to integrate since the first week of the term.
Why an odd exponent is the thing that matters
Suppose $m$ is odd. Then $m-1$ is even, so $\sin^{m}x=\left(\sin^{2}x\right)^{(m-1)/2}\sin x$, and the bracket is a whole number of copies of $\sin^{2}x$.
Each copy can be replaced by $1-\cos^{2}x$, since that is the Pythagorean identity rearranged. So the whole integrand becomes a polynomial in $\cos x$, multiplied by the one $\sin x$ we set aside.
That leftover $\sin x\,dx$ is exactly $-du$ for $u=\cos x$. Nothing of the old variable survives, so the substitution closes.
If $m$ were even, $\sin^{m}x$ would be a polynomial in $\cos x$ with nothing left over to serve as $du$, and there would be no substitution to make. That is the whole content of the parity condition.
The whole decision, in the order you should make it. The only question asked is about $\textcolor{#6f42c1}{\text{parity}}$, never about size: an exponent of $101$ is no harder than an exponent of $3$ as long as it is odd. The $\textcolor{#d1690a}{\text{both even}}$ branch is the one that needs the next concept.
Looks like this, but is not
$\int\sin^{2}x\cos^{2}x\,dx$ is a product of a power of sine and a power of cosine, exactly the shape this method was written for.
Both exponents are even, so peeling a $\sin x$ leaves $\sin^{1}x$ behind: an odd power, which the identity cannot convert into a polynomial in $\cos x$. Nothing has gone wrong with the method; the method simply has no starting move here. That case is the subject of the next concept, and it uses a different identity.
Odd sine, even cosine: ∫ sin³x cos²x dx
The example the naive attempt above stalled on, finished this time.
Given
$$\displaystyle\int\sin^{3}x\cos^{2}x\,dx$$
Find
the antiderivative
Solution
We peel a sine because the sine exponent is the odd one. Peeling a cosine instead would leave $\cos^{1}x$ standing, which the identity cannot convert, so the choice is forced rather than preferred.
Differentiate: $$-\frac{3\cos^{2}x(-\sin x)}{3}+\frac{5\cos^{4}x(-\sin x)}{5}=\cos^{2}x\sin x-\cos^{4}x\sin x=\cos^{2}x\sin x\left(1-\cos^{2}x\right)=\sin^{3}x\cos^{2}x$$. The identity used on the way in is the same one that closes the check on the way out.
One peel, one identity, one substitution, two power-rule integrations.
An odd power on its own: ∫ cos⁵x dx
Only one function appears, so the other exponent is zero. Zero is even, which changes nothing: the method only needs one odd exponent.
Given
$$\displaystyle\int\cos^{5}x\,dx$$
Find
the antiderivative
Solution
Peel a cosine and substitute $u=\sin x$. The alternative, power reduction applied twice, would work but produces $\cos 2x$ and $\cos 4x$ terms and roughly triple the algebra, because power reduction is built for even exponents.
Differentiate: $\cos x-2\sin^{2}x\cos x+\sin^{4}x\cos x=\cos x\left(1-\sin^{2}x\right)^{2}=\cos x\cdot\cos^{4}x=\cos^{5}x$. Recognising $1-2s^{2}+s^{4}$ as $\left(1-s^{2}\right)^{2}$ is what makes the check one line instead of five.
An odd exponent is never a difficulty in this family whatever its size; difficulty here comes from even exponents, not from large ones.
A cotangent inside the product: ∫ cos²x cot x dx
This integrand has a cotangent in it, so there are no exponents to read yet.
Given
$$\displaystyle\int\cos^{2}x\cot x\,dx$$
Find
the antiderivative
Solution
Rewrite first, choose a route second. Every branch in this block is stated for $\sin^{m}x\cos^{n}x$, and $\cot x=\frac{\cos x}{\sin x}$ puts the integrand into exactly that shape for the price of one line.
now the exponents can be read: $3$ on the cosine, and a sine in the denominator, which is an exponent of $-1$. It is left in the denominator rather than written as $\sin^{-1}x$, because that notation is already taken by the inverse function
Read the exponents, then peel the cheaper one
$$m=-1\ \text{is odd},\qquad n=3\ \text{is odd}$$
a negative exponent still has a parity, so a function in the denominator counts as an odd power and both peels are legal in principle
peel the cosine, the one whose exponent is positive, because what is left is a polynomial in $\sin x$ over $\sin x$, which splits term by term; peeling the sine instead leaves $\frac{u^{3}}{1-u^{2}}$ with $u=\cos x$, a rational function that waits for partial fractions in the next block
Substitute and split the fraction
$$u=\sin x,\qquad du=\cos x\,dx$$
the peeled factor is exactly $du$, which is the whole reason for peeling it
A completely different split lands on the same answer with no substitution at all: $\cos^{2}x\cot x=\left(1-\sin^{2}x\right)\cot x=\cot x-\sin x\cos x$, whose two standard pieces integrate to $\ln\lvert\sin x\rvert$ and $\frac{\sin^{2}x}{2}$. Size check as well: on $\left[\frac{\pi}{4},\frac{\pi}{2}\right]$ the integrand falls from $0.5$ to $0$ across a window of width $0.79$, so a straight-line fall would give $\frac{0.5\times 0.79}{2}\approx 0.20$; it is not straight, and at the middle of the window the integrand is already $0.06$, so the value sits well under that. The antiderivative gives $0.097$.
One line of rewriting bought the entire route, and a tangent runs the same way with the two functions exchanged.
Two answers that look different can be the same answer. A key printing $-\ln\lvert\csc x\rvert$ has written our $\ln\lvert\sin x\rvert$, because $\csc x=\frac{1}{\sin x}$; the same pairing turns $-\ln\lvert\cos x\rvert$ into the $\ln\lvert\sec x\rvert$ that $\int\tan x\,dx$ is usually printed as. Check the reciprocal before deciding you are wrong.
Checkpoint
§13.5 — choosing the substitution from the parities●●○○○
One integrand, and only the first move is being asked for. No integration is needed to answer.
Given
$$\displaystyle\int\sin^{4}x\cos^{3}x\,dx$$
Find
Which factor do you peel, and what is $u$?
Hint 1/4
Look only at the two exponents and ask which one is odd. Nothing else about the integrand matters yet.
Hint 2/4
You peel a factor of the function with the odd exponent, and you substitute for the other function.
Hint 3/4
Here $\sin^{4}x\cos^{3}x$ has an even sine exponent and an odd cosine exponent, so the peeled factor is a $\cos x$ and the substitution is for the sine.
Hint 4/4
Peel one $\cos x$ and take $u=\sin x$, with $du=\cos x\,dx$.
Show solution
We check the cosine exponent first only because it happens to be the odd one here; had both been odd and positive, either route would close and the cheaper one would be the one with the smaller exponent.
Differentiate the finished antiderivative: $\sin^{4}x\cos x-\sin^{6}x\cos x=\sin^{4}x\cos x\left(1-\sin^{2}x\right)=\sin^{4}x\cos^{3}x$, the original integrand.
⚠ Substituting without converting what is left
the substitution looks complete as soon as a $du$ has been found, and the surviving even power is easy to overlook
the two substitutions look symmetric, and they are, except that one of the two derivatives carries a minus
wrong$$u=\cos x\ \Rightarrow\ \sin x\,dx=du$$
right$$u=\cos x\ \Rightarrow\ \sin x\,dx=-du$$
⚠ Peeling the larger of two odd positive exponents
the eye goes to the bigger power first, and both routes are legal, so nothing signals that the work has just doubled; with a negative exponent the rule reverses
13.6When both powers are even: turning products into sums
With no odd exponent to peel, halve the powers using $\sin^2x=(1-\cos 2x)/2$ and turn products of angles into sums.
The peel needs an odd exponent, and $\int\sin^{2}x\cos^{2}x\,dx$ has none. What is left is to change the integrand before integrating it.
RuleIdentities that replace a product by a sum
Conditions
the power reduction identities apply to any even power, and are used repeatedly when the exponent is $4$ or more
each application doubles the angle, so keep track of whether you are looking at $x$, $2x$ or $4x$
the last three lines, the product-to-sum identities, are for two different angles multiplied together with no power on either of them; the grouping $\sin x\cos x=\frac{\sin 2x}{2}$ on the first line is a different animal and is listed here only because it is what shortens an even case such as $\sin^{2}x\cos^{2}x$. Both of its exponents are odd, so on its own $\int\sin x\cos x\,dx$ belongs to the peel, which finishes it in one line with $u=\sin x$
These identities are the last step and not the first: an angle that has been shifted or multiplied has to be opened first with the addition formulas $\sin(A\pm B)$ and $\cos(A\pm B)$, and the double angle read in its factored form $\cos 2x=(\cos x-\sin x)(\cos x+\sin x)$, which is what makes a $\cos 2x$ in a numerator the derivative of a $\sin x+\cos x$ in a denominator. The product-to-sum restriction above limits that line to different angles with no power involved, and this case carries a power and a shifted angle at once.
two different angles carrying even powers, as in $\sin^{2}3x-\cos^{2}\frac{x}{2}$, belong to neither of those two lines: reduce each term at its own angle and leave them apart, since an integral of a sum is a sum of integrals. The angle you reduce to is $2ax$, so a half angle comes back as $x$ and an angle of $\pi x$ comes back as $2\pi x$
Every line here does the same job: it takes something multiplied and hands back something added. That matters because a sum can be integrated term by term while a product cannot. The price is that the angles change, so a $\cos 2x$ integrates to $\frac{1}{2}\sin 2x$ and a $\cos 4x$ to $\frac{1}{4}\sin 4x$, and those fractions are the part people forget.
Where the power reduction identities come from
The double-angle identity for cosine has three equivalent forms, and two of them are the ones we want.
From $\cos 2x=1-2\sin^{2}x$, solving for $\sin^{2}x$ gives $\sin^{2}x=\frac{1-\cos 2x}{2}$.
From $\cos 2x=2\cos^{2}x-1$, solving for $\cos^{2}x$ gives $\cos^{2}x=\frac{1+\cos 2x}{2}$.
So nothing new is being introduced. The identity you already know is being read in the other direction: instead of using it to simplify $\cos 2x$, we use it to remove a square. The plus and minus signs are worth a moment: cosine squared, the larger of the two on average near $x=0$, is the one that gets the plus.
Why $\int\cos^{2}x\,dx$ has to come out as $\textcolor{#6f42c1}{\frac{x}{2}}$ plus a wave. The curve $\textcolor{#1f6feb}{y=\cos^{2}x}$ never goes negative and never exceeds $1$, and it spends exactly as much of its time above the level $\textcolor{#6f42c1}{\frac{1}{2}}$ as below it. The identity $\cos^{2}x=\frac{1}{2}+\frac{1}{2}\cos 2x$ says the same thing in symbols: a constant height of one half, plus a wave that averages to nothing.
Looks like this, but is not
$\int\cos^{2}x\,dx$ has a power in it, so the power rule looks applicable: $\frac{\cos^{3}x}{3}$.
Differentiate that and see: $\frac{d}{dx}\frac{\cos^{3}x}{3}=\cos^{2}x\cdot(-\sin x)$, which carries a $\sin x$ the integrand never had. The power rule for integration works on $x$ raised to a power, not on a function raised to a power; the missing piece is exactly the derivative of the inside. The figure gives a second reason to be suspicious before differentiating anything: $\int_{0}^{2\pi}\cos^{2}x\,dx$ is visibly a positive area, about $\pi$, while $\frac{\cos^{3}x}{3}$ returns to its starting value after a full period and would predict zero.
The basic even case: ∫ cos²x dx
One identity, one integration, and one factor of $\frac{1}{2}$ that has to survive the trip.
Given
$$\displaystyle\int\cos^{2}x\,dx$$
Find
the antiderivative
Solution
Power reduction rather than parts. Parts does work here and produces the same answer through a , but it costs two passes and an equation to solve, where the identity costs one line.
Replace the square by a sum
$$\cos^{2}x=\frac{1+\cos 2x}{2}$$
the identity, chosen with a plus sign because cosine squared is $1$ at $x=0$ while the right side must also be $1$ there
Differentiate: $\frac{1}{2}+\frac{2\cos 2x}{4}=\frac{1+\cos 2x}{2}=\cos^{2}x$. Independently, over one full period the wave term returns to where it started, so $\int_{0}^{2\pi}\cos^{2}x\,dx=\pi$, which is the average height $\frac{1}{2}$ times the length $2\pi$ that the figure predicted.
An even power of sine or cosine always integrates to a multiple of $x$ plus waves, so an answer with no $x$ in it has lost something.
Both exponents even: ∫ sin²x cos²x dx
Two even powers. Applying the two identities separately works but creates a product of two brackets; there is a shorter road.
Given
$$\displaystyle\int\sin^{2}x\cos^{2}x\,dx$$
Find
the antiderivative
Solution
We use $\sin x\cos x=\frac{1}{2}\sin 2x$ first rather than expanding $\frac{1-\cos 2x}{2}\cdot\frac{1+\cos 2x}{2}$, because grouping the product before squaring turns two identities into one and removes the cross terms entirely.
the two constants $\frac{1}{4}$ and $\frac{1}{2}$ combine into $\frac{1}{8}$
$$=\frac{x}{8}-\frac{\sin 4x}{32}+C$$
the angle $4x$ contributes a further $\frac{1}{4}$, which is where the $32$ comes from
Answer $$\frac{x}{8}-\frac{\sin 4x}{32}+C$$
Check
Differentiate: $\frac{1}{8}-\frac{4\cos 4x}{32}=\frac{1-\cos 4x}{8}$, and running the two identities forwards turns that back into $\frac{\sin^{2}2x}{4}=\sin^{2}x\cos^{2}x$. A second check on size: the integrand never exceeds $\frac{1}{4}$, so over $\left[0,2\pi\right]$ the answer must be below $\frac{2\pi}{4}\approx 1.57$, and it is $\frac{2\pi}{8}\approx 0.785$.
Two identities and one integration. Expanding the brackets instead costs an extra $\cos^{2}2x$ that has to be reduced anyway.
Different angles, no powers: ∫ sin 5x cos 3x dx
Nothing here is squared, so neither peeling nor power reduction has anything to bite on. The obstacle is that the two angles are different.
Given
$$\displaystyle\int\sin 5x\cos 3x\,dx$$
Find
the antiderivative
Solution
The product-to-sum identity, because a substitution cannot work: neither factor is a constant multiple of the derivative of anything inside the other, and parts would return the integral to itself after two passes with coefficient $1$, which says nothing.
Differentiate: $\frac{8\sin 8x}{16}+\frac{2\sin 2x}{4}=\frac{\sin 8x+\sin 2x}{2}$, which is the identity read backwards, so it is $\sin 5x\cos 3x$. As a numerical spot check at $x=\frac{\pi}{4}$: the integrand is $\sin\frac{5\pi}{4}\cos\frac{3\pi}{4}=\left(-\frac{\sqrt2}{2}\right)\left(-\frac{\sqrt2}{2}\right)=\frac{1}{2}$, and the derivative formula gives $\frac{\sin 2\pi+\sin\frac{\pi}{2}}{2}=\frac{0+1}{2}=\frac{1}{2}$.
Whenever two different angles are multiplied, reach for these identities first; the same three lines handle every product of that shape.
Two different angles, both squared: ∫ (6sin²(x/2) − cos²2x) dx
Two even powers, at two angles, and neither angle is $x$. Both branches on the route card look wrong for this, and one line of reading fixes that.
Power reduction, applied to each term on its own. The peel needs an odd exponent and there is none; the product-to-sum line needs two functions multiplied, and here they are subtracted, each carrying its own square. Nothing has to be combined first, because an integral of a sum is a sum of integrals and the two terms never meet.
Average heights over one full period $\left[0,2\pi\right]$: $\sin^{2}\frac{x}{2}$ averages $\frac{1}{2}$, so the first term contributes $6\cdot\frac{1}{2}\cdot 2\pi=6\pi$, and $\cos^{2}2x$ also averages $\frac{1}{2}$, contributing $\pi$; the difference is $5\pi$. The antiderivative gives $\frac{5}{2}\cdot 2\pi-0-0=5\pi$. Differentiating is the second, independent check: $\frac{5}{2}-3\cos x-\frac{4\cos 4x}{8}$ folds straight back into the integrand.
Two identities, no substitution, no product to expand. Anything that combined the two terms first would have been longer.
Two habits to carry out of this. The reduced angle is $2ax$ and not $2x$, so read $a$ off the inside of the function every time; and terms that are added stay apart, so a mixture of angles is not an obstacle. Both traps arrive together the moment a volume integrand such as $\cos^{2}(\pi x)$ shows up, where copying the $a=1$ formula quietly loses a factor of $\pi$.
Checkpoint
§13.6 — integrating an even power●●○○○
A one-line integral that appears somewhere in almost every question of this family. The answer is being asked for directly, not the method.
Given
$$\displaystyle\int\sin^{2}x\,dx$$
Find
Which one is the antiderivative?
Hint 1/4
There is no odd factor to peel, so the integrand has to be rewritten before it can be integrated at all.
Hint 2/4
The identity is $\sin^{2}x=\frac{1-\cos 2x}{2}$, and the sign is fixed by checking $x=0$, where both sides must be $0$.
Hint 3/4
That gives $\frac{1}{2}\int\left(1-\cos 2x\right)dx$, and $\int\cos 2x\,dx=\frac{\sin 2x}{2}$ because the angle is $2x$.
Hint 4/4
The antiderivative is $\frac{x}{2}-\frac{\sin 2x}{4}+C$.
Show solution
Power reduction rather than the Pythagorean identity: writing $\sin^{2}x=1-\cos^{2}x$ would leave $\int\cos^{2}x\,dx$, an integral of exactly the same difficulty, so nothing would have been gained.
Rewrite, then integrate
$$\sin^{2}x=\frac{1-\cos 2x}{2}$$
checked at $x=0$: both sides give $0$, which fixes the sign
the inner angle $2x$ produces the extra $\frac{1}{2}$
$$=\frac{x}{2}-\frac{\sin 2x}{4}+C$$
collecting
Answer $$\frac{x}{2}-\frac{\sin 2x}{4}+C$$
Check
Add this to the answer for $\int\cos^{2}x\,dx$ found above: the sine terms cancel and the sum is $x+C$, which is what $\int\left(\sin^{2}x+\cos^{2}x\right)dx=\int 1\,dx$ must give. That check uses neither of the two computations again.
⚠ Applying the power rule to a squared trigonometric function
$\cos^{2}x$ looks like a power, and the power rule is the first tool anyone reaches for when they see one
⚠ Forgetting the reciprocal factor from the inner angle
the integrand is written in terms of $2x$ or $4x$ but the antiderivative is written by reflex as if the angle were $x$
wrong$$\int\cos 4x\,dx=\sin 4x+C$$
right$$\int\cos 4x\,dx=\frac{\sin 4x}{4}+C$$
⚠ Getting the sign of the backwards
the two identities differ by one symbol and are memorised together, so they are easy to swap under time pressure
wrong$$\sin^{2}x=\frac{1+\cos 2x}{2}$$
right$$\sin^{2}x=\frac{1-\cos 2x}{2}$$
⚠ Reducing a scaled angle as though the angle were x
the identity is memorised in its $a=1$ form, and once it has been written down the coefficient inside the function is never looked at again
wrong$$\cos^{2}(\pi x)=\frac{1+\cos 2x}{2}$$
right$$\cos^{2}(\pi x)=\frac{1+\cos 2\pi x}{2}$$
13.7Integrals with no right-hand end
An infinite endpoint or a blow-up means an ordinary integral inside a limit; converge or diverge is part of the answer.
Every integral so far ran between two numbers, over a function that stayed finite. This week's syllabus also lists the case where one of those fails.
Solvable with what we have
$\int_{1}^{5}\frac{dx}{x^{2}}$ — a bounded interval and a bounded integrand, so the Fundamental Theorem applies
$\int_{0}^{\pi}x\sin x\,dx$ — parts, then evaluate at two numbers
$\int_{1}^{2}\frac{dx}{x}$ — a logarithm, evaluated at two numbers
Not solvable yet
$\int_{1}^{\infty}\frac{dx}{x^{2}}$
$\int_{0}^{1}\frac{dx}{\sqrt{x}}$
$\int_{0}^{\infty}xe^{-x}\,dx$
Treat $\infty$ as if it were a number and substitute it: $\int_{1}^{\infty}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{1}^{\infty}=-\frac{1}{\infty}+1=1$. The number is right, which is the dangerous part. Run the same reasoning on $\int_{-1}^{1}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{-1}^{1}=-1-1=-2$, and a strictly positive integrand has produced a negative area.
Why it fails
The Fundamental Theorem asks for a continuous function on a closed bounded interval, and neither example supplies one. Substituting $\infty$ works by accident whenever the limit happens to exist, and gives a confident wrong answer whenever it does not. Writing the limit down every time is what makes the second case visible.
DefinitionImproper integrals of Type I and Type II
Conditions
$f$ is continuous on the interval with the bad endpoint removed
if both endpoints are bad, or a blow-up sits strictly inside the interval, the integral is split at a convenient interior point and every piece must converge on its own
convergence means the defining limit exists and is finite; anything else, including a limit that runs off or oscillates, is divergence
Convergence is decided by the bad end alone: an inequality, or a limit of a ratio, that holds only on $[X,\infty)$ for some $X\ge a$, or only on $(a,a+\delta]$, settles the whole question, because whatever is left over is an ordinary definite integral of a continuous function on a closed bounded interval and is finite whatever its size.
Before splitting anything, check whether the suspect point is bad at all. If the integrand has a finite limit there, the singularity is removable and the integral over a finite interval is ordinary; if the integrand merely stays bounded there, however violently it oscillates, the integral is again ordinary. Neither case needs a comparison, and neither is a candidate for one.
One integral can be improper in two different ways at the same time, for instance a blow-up at the finite left endpoint together with an infinite tail. It is then split at a convenient interior point and each regime is given its own comparator, since the power that describes the integrand near the singularity is never the one that describes it far out.
An improper integral is not a new kind of integral: it is a limit of the ones you already know, taken over a growing or shrinking window. Do the ordinary definite integral with a letter in place of the bad endpoint, then let the letter move. The last line is the reference family, and in both halves of it the exponent $1$ is the dividing line that belongs to neither side.
Why the exponent 1 is the dividing line
Compute the window integral first, keeping $t$ as a letter. For $p\neq 1$, $\int_{1}^{t}x^{-p}\,dx=\frac{t^{1-p}-1}{1-p}$; for $p=1$ it is $\ln t$.
Now let $t\to\infty$. The behaviour of $t^{1-p}$ decides everything: if $p>1$ the exponent $1-p$ is negative, so $t^{1-p}\to 0$ and the whole expression tends to $\frac{1}{p-1}$. If $p<1$ the exponent is positive and $t^{1-p}\to\infty$, so the integral diverges.
The case $p=1$ sits between them and has to be checked separately, because the antiderivative changes shape: $\ln t\to\infty$, slowly, but with no ceiling. Slow growth is still growth, which is why $p=1$ falls on the divergent side.
Near zero the same computation runs with the roles of the two ends exchanged, and the inequality flips.
Two integrands that look alike and behave oppositely. Both start at height $1$ and both fall to zero, but the area collected under $\textcolor{#d1690a}{y=\frac{1}{x}}$ is $\ln t$, which has no ceiling, while the area under $\textcolor{#1f6feb}{y=\frac{1}{x^{2}}}$ is $1-\frac{1}{t}$ and stops at $1$. Decaying to zero is not enough; the question is how fast.
Looks like this, but is not
$\int_{-1}^{1}\frac{dx}{x^{2}}$ has two perfectly ordinary numbers as its limits, so it looks like an ordinary definite integral.
The integrand blows up at $x=0$, which is strictly inside the interval, so the integral is improper even though nothing in the notation says so. It must be split as $\int_{-1}^{0}+\int_{0}^{1}$, each piece taken as its own limit, and both diverge. A negative number from a positive integrand is the tell that the Fundamental Theorem was used where it does not hold.
$p$
$t=10$
$t=100$
$t=10^{4}$
as $t\to\infty$
$\tfrac{1}{2}$
$4.32$
$18.00$
$198.00$
diverges, like $2\sqrt{t}$
$1$
$2.30$
$4.61$
$9.21$
diverges, like $\ln t$
$2$
$0.90$
$0.99$
$0.9999$
converges to $1$
The middle row is the one worth staring at. Multiplying the window by $100$ adds only about $4.6$ to the total, so the growth is very slow, and a table of numbers alone could easily be read as settling down. It does not settle down: $\ln t$ passes every ceiling eventually. This is why the verdict comes from the limit and not from a table.
The reference family: ∫₁^∞ dx/xᵖ
One computation that answers a whole class of later questions, done once with the exponent left as a letter.
Given
$$\displaystyle\int_{1}^{\infty}\frac{dx}{x^{p}}$$, with $p>0$ a constant
Find
the values of $p$ for which it converges, and the value when it does
Solution
We keep $p$ as a letter instead of testing values, because the whole point of this family is to be used as a reference later; a single computation with a letter in it replaces a dozen with numbers.
Test the formula at a value computed independently: $p=2$ gives $\frac{1}{2-1}=1$, and directly $\int_{1}^{t}x^{-2}\,dx=1-\frac{1}{t}\to 1$. Also check the boundary behaviour: as $p$ decreases towards $1$ the value $\frac{1}{p-1}$ grows without bound, which is what it must do if the case $p=1$ is divergent.
Far out to the right a large exponent is your friend, because it makes the decay fast, and near a blow-up at zero it is the enemy, because it makes the spike tall.
A blow-up at an endpoint: ∫₀¹ dx/√x
The interval is as ordinary as it looks. The integrand is not: it runs off to infinity as $x$ approaches the left endpoint.
Given
$$\displaystyle\int_{0}^{1}\frac{dx}{\sqrt{x}}$$
Find
whether it converges, and its value if it does
Solution
The limit is taken at the left endpoint because that is where the integrand is unbounded; the right endpoint is perfectly ordinary and no limit is needed there.
the power rule with exponent $-\frac{1}{2}$, which is greater than $-1$, so no logarithm appears
$$\lim_{t\to 0^{+}}\left(2-2\sqrt{t}\right)=2$$
the square root is continuous at $0$, so the limit is an ordinary substitution once the integral has been done
Answer $$\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$$
Check
Check against the reference family: this is $\int_{0}^{1}x^{-p}\,dx$ with $p=\frac{1}{2}<1$, which the boxed rule says converges, and to $\frac{1}{1-p}=2$.
An unbounded region can have finite area: what matters is how fast the height grows against how fast the width shrinks.
Improper at both ends: ∫ dx/(x²+4) over the whole line
Neither endpoint is a number, so neither one can be substituted, and no single letter can chase both of them.
Split the interval and run two independent limits, rather than letting one letter grow in both directions at once. The symmetric window $\int_{-t}^{t}$ is tempting because this integrand is even, but it is not the definition, and on a different integrand it manufactures a value where none exists.
Split, because one limit cannot chase two endpoints
A size bound that never touches the antiderivative: for $\lvert x\rvert\ge 2$ the integrand is at most $\frac{1}{x^{2}}$, so the two tails contribute at most $2\int_{2}^{\infty}\frac{dx}{x^{2}}=1$, and on $\left[-2,2\right]$ the integrand is at most $\frac{1}{4}$ over a length of $4$, contributing at most $1$. The total therefore sits below $2$, and $\frac{\pi}{2}\approx 1.57$ does.
The split is not bookkeeping. Run the same set-up on $\int_{-\infty}^{\infty}\frac{x\,dx}{1+x^{2}}$: each half is $\frac{1}{2}\ln\left(1+x^{2}\right)$ measured against an infinite end, so each half diverges and the whole integral diverges, while the symmetric window is $0$ for every $t$ by symmetry and would have reported $0$ with a straight face. One divergent half ends the question.
Checkpoint
§13.7 — reading a verdict off the exponent●●○○○
One tail integral, and the reference family above is enough to settle it without computing anything from scratch.
You do not need to integrate. This is a member of the reference family, so identify the exponent and compare it with the dividing line.
Hint 2/4
$\int_{1}^{\infty}x^{-p}\,dx$ converges exactly when $p>1$, and its value is then $\frac{1}{p-1}$.
Hint 3/4
Here $p=\frac{3}{2}$, so $p>1$ and $p-1=\frac{1}{2}$.
Hint 4/4
It converges, and the value is $\frac{1}{1/2}=2$.
Show solution
The reference family rather than a fresh limit computation, because the general case was already done with a letter and this is that letter set to $\frac{3}{2}$.
Identify p and apply the rule
$$p=\frac{3}{2}>1$$
the tail integral converges exactly on this side of the line
$$\frac{1}{p-1}=\frac{1}{\tfrac{1}{2}}=2$$
the value from the boxed formula
Answer $$\int_{1}^{\infty}\frac{dx}{x^{3/2}}=2$$
Check
Do the limit directly as a check: $\int_{1}^{t}x^{-3/2}\,dx=\left[-2x^{-1/2}\right]_{1}^{t}=2-\frac{2}{\sqrt{t}}\to 2$. Two independent routes, one answer.
⚠ Substituting infinity into the antiderivative
the notation puts $\infty$ exactly where a number usually sits, and the answer often comes out right, so the habit survives unpunished until it does not
A product of two factors with no substitution available, or a single factor whose derivative is simpler than itself.
Rule out substitution first
Ask whether one factor is a constant multiple of the derivative of something inside the other. If it is, stop; substitution is cheaper.
Choose u by the order
Logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. The earliest class present becomes $u$, and everything left, $dx$ included, is $dv$.
Write all four pieces on one line
$u$, $du$, $dv$, $v$, before touching the integral. Most sign errors are born in the gap between choosing and substituting.
Substitute into the formula with brackets
$uv-\int v\,du$, with the minus sign written as if it covered a bracket, because it does.
Judge the trade
Look at $\int v\,du$. Easier means carry on; the same difficulty means you may be in the returning case; harder means swap the roles and restart.
Where it goes wrong
$dv$ chosen so that $v$ cannot be written down at all
the same factor differentiated on one pass and antidifferentiated on the next, which returns you to the start
the minus sign applied to the first term of $\int v\,du$ only
The table, for a polynomial against something that never dies
$\int p(x)f(x)\,dx$ where $p$ is a polynomial and $f$ is $e^{ax}$, $\sin ax$ or $\cos ax$.
Two columns
Left column: $p$ and its derivatives. Right column: $f$ and its repeated antiderivatives.
Walk the left column to zero
A polynomial of degree $n$ needs $n+1$ rows to reach $0$.
Match the right column row for row
Each antidifferentiation of $e^{ax}$ multiplies by $\frac{1}{a}$, and of $\sin ax$ or $\cos ax$ by $\frac{1}{a}$ with a sign change every second step.
Multiply diagonally
Row $k$ on the left against row $k+1$ on the right.
Alternate the signs, starting with a plus
Add the signed products and attach $+C$.
Where it goes wrong
starting the sign pattern with a minus, which negates the whole answer
using the table when the polynomial belongs on the $dv$ side, as in $\int x^{2}\ln x\,dx$
forgetting the factor $\frac{1}{a}$ on each antidifferentiation, which corrupts every row below the first
When the integral you started with comes back
$\int e^{ax}\sin bx\,dx$, $\int e^{ax}\cos bx\,dx$, $\int\sec^{3}x\,dx$ and anything else where neither factor simplifies.
Name the integral
Write $I=\int\ldots$ at the top. You are about to solve for it, and it needs a name.
Two passes, same orientation
Keep the same class of function as $u$ on both passes. Swapping halfway undoes the first pass exactly.
Recognise the return
The line now reads $I=A-kI$ or $I=A+kI$, where $A$ has no integral in it.
Solve, then restore the constant
Collect $I$ on one side and divide. If the coefficient of $I$ works out to $1$ on the wrong side, the two passes cancelled and you must redo the second one.
Where it goes wrong
swapping $u$ and $dv$ on the second pass and arriving at $I=I$
treating the return as evidence that the method has failed
leaving out $+C$ after dividing, because $I$ looked like a number while the algebra was being done
Choosing the route for a product of sines and cosines
$\int\sin^{m}x\cos^{n}x\,dx$, and products of sines and cosines of different angles, and any product carrying a tangent, cotangent, secant or cosecant once it has been rewritten in sines and cosines. A sum of such terms is split first and routed one term at a time.
First make the integrand sines and cosines only
Nothing below can fire while a $\tan$, $\cot$, $\sec$ or $\csc$ is standing there, because every branch here is written for $\sin^{m}x\cos^{n}x$. Replace them by $\frac{\sin x}{\cos x}$, $\frac{\cos x}{\sin x}$, $\frac{1}{\cos x}$, $\frac{1}{\sin x}$ and then read the exponents: a function in the denominator is a negative exponent, and $-1$ is odd, so $\int\sin^{2}x\tan x\,dx$ is an odd case rather than a case with no route. The family to leave alone is the tangent and secant one: every $\int\tan^{m}x\sec^{n}x\,dx$, a lone $\int\sec^{3}x\,dx$ or $\int\tan^{2}x\,dx$ included, has its own box below, as does its cotangent and cosecant mirror, and rewriting them here walks into partial fractions or into a dead end at branch 5. Rewrite when the tangent or cotangent stands next to a power of a sine or a cosine, which is what leaves an exponent to peel.
Read the two exponents
Only their parity matters. Size does not.
One of them odd
Peel a factor of the function with the odd exponent, convert what remains with $\sin^{2}x+\cos^{2}x=1$, and substitute for the other function.
Both odd
Either route closes, but the tie-break depends on the signs. With both exponents positive, peel the smaller one, because the surviving bracket then has a lower power to expand. With one of them negative — the $-1$ a rewritten $\tan$ or $\cot$ leaves in a denominator — peel the positive one instead, since peeling out of a denominator leaves a rational function of $u$, which is the next block.
Both even
No peel exists. Use $\sin^{2}x=\frac{1-\cos 2x}{2}$ and $\cos^{2}x=\frac{1+\cos 2x}{2}$, and group $\sin x\cos x$ into $\frac{\sin 2x}{2}$ first when both appear.
Different angles instead of powers
Use the product-to-sum identities; there is nothing to peel and nothing to reduce.
Different angles carrying powers, and sums
A sum is not a product, so split it before routing anything: $\int(f+g)\,dx=\int f\,dx+\int g\,dx$, and each term then meets the branches above on its own. That is the way in when two different angles each carry an even power and are added, as in $\cos^{2}\frac{x}{2}+\sin^{2}3x$: as one object it matches nothing here, since branch 5 is written at the angle $x$ and branch 6 wants a product with no powers. Reduce each term at its own angle, reading $a$ off the inside of the function, so $\sin^{2}(ax)=\frac{1-\cos 2ax}{2}$. Multiplied rather than added, as in $\sin^{2}3x\cos^{2}\frac{x}{2}$, there is no sum to split: reduce each factor at its own angle anyway, and the product of two cosines left over is branch 6.
Where it goes wrong
substituting before converting the surviving even power, so the old variable never leaves
peeling the larger of two odd exponents while both are positive; and the mirror slip, peeling the negative one and landing in partial fractions
reaching for power reduction when an odd exponent was available
trying to match a $\tan$ or a $\cot$ against the $\sin^{m}x\cos^{n}x$ pattern instead of rewriting it first, and concluding from the failure that the integral needs a technique nobody has taught
reading a sum of terms at different angles as one object and hunting for the branch that swallows it whole, when each term is routed on its own
The tangent and secant family, in one paragraph
$\int\tan^{m}x\sec^{n}x\,dx$, the companion family to the sine and cosine one.
The identity to use
$\sec^{2}x=1+\tan^{2}x$ plays the role that $\sin^{2}+\cos^{2}=1$ played before.
Even power of secant
Peel $\sec^{2}x$ as the differential of $\tan x$, convert the rest with the identity, and take $u=\tan x$.
Odd power of tangent
Peel $\sec x\tan x$ as the differential of $\sec x$, convert the rest, and take $u=\sec x$.
Neither, as in an odd power of secant alone
You are in the returning case: $\int\sec^{3}x\,dx$ is done by parts and solved for, not by an identity.
The two antiderivatives worth knowing
$\int\tan x\,dx=\ln\lvert\sec x\rvert+C$ and $\int\sec x\,dx=\ln\lvert\sec x+\tan x\rvert+C$.
Where it goes wrong
trying to peel when the powers are $\tan^{2}$ against $\sec^{0}$, where the identity alone finishes the job
forgetting that $\int\sec^{3}x\,dx$ needs parts rather than a substitution
dropping the absolute value bars in the two logarithmic antiderivatives
Setting up an integral with a bad endpoint
Any integral with $\infty$ as a limit, or whose integrand is unbounded somewhere on the interval.
Find every bad point
Infinite limits, and points where the integrand is undefined or unbounded. Check inside the interval as well as at its ends.
Split so that each piece has at most one bad point, at an end
A blow-up inside the interval must be split at, so the two pieces each carry it as an endpoint. The same move covers an integral that is bad at both ends: two infinite limits, or an infinite limit at one end and a blow-up at the other. Split at any convenient interior point, give each piece its own letter, and require both to converge.
Replace the bad endpoint by a letter
Write $\lim_{t\to\ldots}$ in front, and from that point on you are doing an ordinary definite integral.
Integrate with the letter still in place
This is where the techniques of this week are usually needed; the limit is taken only after the antiderivative has been evaluated.
Take the limit and report a verdict
Finite means converges, with the number attached. Anything else means diverges, and every piece of a split integral must converge for the whole to converge.
Where it goes wrong
substituting $\infty$ into the antiderivative instead of taking a limit
splitting at the bad point but then declaring convergence because one of the two pieces converges
taking the limit before doing the integral
letting one letter run to both ends at once, as in $\lim_{t\to\infty}\int_{-t}^{t}$, which is not the definition and reports a number for integrals that diverge
Deciding an improper integral when there is no antiderivative
The question says only "determine whether the integral converges" and asks for no value, or the integrand has no antiderivative worth chasing. This is the most frequent shape in the whole pool for this week: the official exercise set closes with five of them and the quiz convergence block is entirely of this kind, and none of them pays anything for an attempted antiderivative.
Ask first whether it is improper at all
List every suspect point: infinite endpoints, and points where the integrand is undefined or unbounded. A suspect point where the integrand has a finite limit is removable and the integral there is ordinary. So is a point where the integrand stays bounded, however violently it oscillates. Only an infinite endpoint or a genuine blow-up survives this step, and a question that looks improper and is not is a question you finish in one sentence.
Split until each piece carries exactly one bad point, at an end
An integral bad at both ends, or bad in two different ways at once, such as a blow-up at the left endpoint together with an infinite tail, is split at any convenient interior point. Each piece is now a separate question with its own reference integral, and the whole converges only if every piece does.
Check the sign before reaching for a comparison
Comparison needs $0\le f$ on the interval where you use it. If the integrand changes sign infinitely often out in the tail, no comparison proves anything about it and the integral has to be handled another way, usually by parts.
Read the leading behaviour and name a candidate
Near a finite bad point $x=a$, factor out the piece that vanishes and read the power of $(x-a)$ that survives. In a tail, drop everything but the dominant term and read the power of $x$. This produces the guess and the comparator, not the proof.
Turn the guess into an inequality on a stated interval
Write $0\le f\le g$ or $0\le g\le f$ and say where it holds. Near the bad point is enough: "for $x\ge X$" or "for $a<x<a+\delta$" settles the question, because the leftover piece is an ordinary integral of a continuous function. Elementary bounds such as $\ln x\le x-1$ or $\lvert\sin\theta\rvert\le 1$ are what convert a non-power integrand into a power comparator. If no inequality is convenient, compute $\lim f/g$ instead and use the Limit Comparison Test.
State the comparator's verdict as its own sentence
Write the reference integral with its integral sign and both endpoints, then say that it converges or diverges. "$\int_{1}^{\infty}\frac{dx}{x^{2}}$ converges" scores; "$\frac{1}{x^{2}}$ converges" does not, and neither does the same claim buried inside a line of algebra.
Write one conclusion sentence that names the test
The given integral converges or diverges because the inequality holds on that interval, and because the reference integral converges or diverges, by the Direct Comparison Test or the Limit Comparison Test. All three clauses go in the one sentence; the examiner's note in the pool says there is no flexible reading of this line.
Where it goes wrong
Bounding above by a divergent integral, or below by a one: every inequality written can be true while the argument excludes nothing
Writing "by comparison" with no comparator, no interval and no verdict for the comparator, which the pool's key gives nothing for
Hunting for an antiderivative on a question that asked only for a verdict
Comparing an integrand that changes sign infinitely often, where the test has no hypothesis to stand on
Declaring the whole integral convergent after settling only one of the pieces it was split into
Using a value computed in a later part of the same question as the reason for convergence in an earlier one, which the key scores at zero
Comparing a blow-up at $x=a$ with a power of $x$ instead of a power of $x-a$
Changing variable inside an improper integral
A named substitution is handed to you for an integral with a bad endpoint, or the antiderivative is only reachable through a substitution. Both finals in the pool end their longest question this way, and the marks are split between the substitution and the limit rather than sitting on the final number.
Replace the bad endpoint by a letter first
Write $\lim_{t\to\ldots}\int_{a}^{t}$ before anything else. From this line on you are doing an ordinary definite integral, and the letter behaves exactly like any other endpoint.
Push both endpoints through the substitution formula, the letter included
If $y=\sqrt{x}$ sends $x=4$ to $y=2$, it sends $x=t$ to $y=\sqrt{t}$. If $y=-\ln x$ sends $x=1$ to $y=0$, it sends $x=a$ to $y=-\ln a$, and a finite bad endpoint has just become an infinite one. Copying the old endpoints onto the new integral is a separately marked mistake.
Convert $dx$ completely, and let the sign fix the order of the ends
A substitution that leaves any of the old variable standing has not been carried out. Where the substitution reverses which end is larger, the minus sign coming out of $dx$ is what puts the two ends back in order.
Integrate in the new variable
The new integrand is what the substitution was chosen to produce, and in this week's questions it is almost always a polynomial against an exponential, which is the table case of integration by parts.
Evaluate at the two new endpoints, and only then take the limit
The boundary term is a function of the letter, not a number. Its limit is a separate step, it is usually indeterminate on first sight, and it is settled either by l'Hospital's rule or by quoting a growth limit that an earlier part of the same question already proved.
Report the verdict together with the value
"Converges to $L$", with the limit written out at least once on the page. The examiner's note in the pool says the emphasis is on the improperness and not on whether you can integrate, so a correct value with no limit anywhere collects about half of what the part is worth.
Where it goes wrong
Carrying the old endpoints onto the new integral after the change of variable
Taking the limit before there is an antiderivative to take it of
Treating $\bigl[uv\bigr]_{a}^{t}$ as a number while one of its ends is still a letter
Re-deriving a growth limit that an earlier part already established, or quoting one that was never established
Producing the correct number with no limit written anywhere on the page
Answering the convergence part by pointing at the value found in the later part, when the two are marked as separate jobs
Substitution finishes it: ∫ x e^{x²} dx
An algebraic factor times an exponential, and the exponent is not $x$.
Given
$$\displaystyle\int xe^{x^{2}}\,dx$$
Find
the antiderivative
SolutionFind the hidden derivative
$$u=x^{2},\qquad du=2x\,dx$$
the exponent is composite and its derivative is present up to the constant $2$
one pass is enough because differentiating $x$ removes it
Answer $$e^{x}\left(x-1\right)+C$$
Check
Differentiate: $e^{x}(x-1)+e^{x}=xe^{x}$.
The two integrands differ by a single exponent, and that one symbol decides whether the $x$ out front is fuel for a substitution or an obstacle that has to be differentiated away.
How to tell them apart
Ask what is inside the outer function and differentiate it. If the result is standing in the integrand up to a constant, substitute. If the result is a constant, there is nothing to absorb the spare factor and you are in a parts problem.
An odd exponent: ∫ sin³x cos³x dx
Both exponents are odd, so either route closes and the cheaper one wins.
both exponents are $3$, so the two routes are exact mirror images and cost the same; we peel the cosine and say so rather than pretending there was a reason
A size check: the integrand never exceeds $\frac{4}{27}\approx 0.148$, so over $\left[0,\pi\right]$ the value must be below $0.47$; the answer gives $\frac{\pi}{16}\approx 0.196$ once the two periodic terms cancel over a full period.
One exponent changes from three to four, and a three-line substitution becomes a two-stage rewrite with two different doubled angles in it.
How to tell them apart
Check parity before anything else, and check it on both exponents. An odd exponent anywhere means peel; only when both are even do you rewrite, and even then a doubled angle can hand you back an odd power to peel.
Scaffolding comes off
The common skeleton
Check for a substitution first: is one factor a constant multiple of the derivative of something inside the other?
Choose $u$ by the order logarithmic, inverse trigonometric, algebraic, trigonometric, exponential; everything else, $dx$ included, is $dv$.
Write $u$, $du$, $dv$ and $v$ down together before substituting anything.
Put them into $uv-\int v\,du$, treating the minus sign as if it covered a bracket.
Look at the new integral: easier means finish it, the same shape means solve for the original, harder means the split was the wrong way round.
Differentiate the answer and compare it with the integrand.
1 · fully worked
Full worked rung: ∫ x e³ˣ dx
Every step and every reason written out. Read it once with the reasons covered, then again with them showing.
Given
$$\displaystyle\int xe^{3x}\,dx$$
Find
the antiderivative
Solution
Parts rather than substitution: the derivative of the exponent $3x$ is the constant $3$, which cannot absorb the spare $x$, so the $x$ has to be differentiated away instead.
Split
$$u=x,\qquad dv=e^{3x}\,dx$$
algebraic beats exponential on the order, and differentiating $x$ terminates while differentiating $e^{3x}$ never does
$$du=dx,\qquad v=\frac{e^{3x}}{3}$$
the linear exponent contributes the factor $\frac{1}{3}$, which will appear again in the next line and must not be dropped
Every antidifferentiation of $e^{ax}$ costs a factor $\frac{1}{a}$, and in a two-pass problem that factor appears twice.
2 · you write the reasoning
Same skeleton, easier numbers: $\int x\sin x\,dx$. The steps are given and the reasons are not. Write your own reason for each line before opening the model answers, because the reasons are the part that transfers to the next problem.
reasoning
Algebraic beats trigonometric on the order, and differentiating $x$ destroys it in one step while differentiating $\sin x$ only turns it into a cosine.
reasoning
One differentiation and one antidifferentiation. The minus in $v$ comes from $\int\sin x\,dx=-\cos x$ and it is the sign that survives into the answer.
reasoning
The formula, written with $v$ kept inside a bracket so that its own minus sign is visible rather than merged into the subtraction in front of it.
reasoning
Two minus signs meeting: the one from the formula and the one inside $v$. Their product is a plus, which is why the final answer adds a sine instead of subtracting one.
reasoning
The traded integral was easier than the original, which is the test that tells you the split was the right way round.
3 · find the buried error
Harder than the last one, because it needs two passes. Below is a student's full solution to $\int x^{2}e^{-x}\,dx$. Exactly two of the five lines contain an error, and each error is one that a careful person makes under time pressure. Find them before looking at the choices.
the two buried errors (2)
⚠ step 2
The minus sign carried by $v$ has been dropped. With $v=-e^{-x}$ and $du=2x\,dx$, the term $-\int v\,du$ is $-\int\left(-e^{-x}\right)2x\,dx=+2\int xe^{-x}\,dx$, so the traded integral enters with a plus, not a minus.
The formula is memorised as "$uv$ minus the integral of the rest", and the sign that $v$ already carries gets absorbed into that memorised minus instead of being multiplied by it.
right
Write $v\,du$ as an explicit product first, $\left(-e^{-x}\right)\left(2x\,dx\right)$, and only then apply the outer minus. The correct line is $\int x^{2}e^{-x}\,dx=-x^{2}e^{-x}+2\int xe^{-x}\,dx$.
⚠ step 3
The final antidifferentiation is wrong: $\int e^{-x}\,dx=-e^{-x}$, not $+e^{-x}$. The line should end at $-xe^{-x}-e^{-x}$.
The same antiderivative was written correctly in step 1, so it feels already checked; the second occurrence gets copied from memory of $e^{x}$ rather than recomputed.
right
Every time $e^{-x}$ is antidifferentiated, write the factor $\frac{1}{-1}$ explicitly. The correct line is $\int xe^{-x}\,dx=-xe^{-x}-e^{-x}$.
4 · the bare problem
§13.3 — the scaffolding is off●●●○○
No steps, no reasons, no split suggested. Two passes are needed and the coefficient in the exponent shows up in every one of them.
Given
$$\displaystyle\int x^{2}e^{2x}\,dx$$
Find
Evaluate the integral, and check your answer by differentiating it.
Hint 1/4
Decide first whether anything can be substituted. The derivative of the exponent is a constant, so the $x^{2}$ has nowhere to go except through differentiation.
Hint 2/4
Use the table: left column $x^{2}$, $2x$, $2$, $0$; right column the repeated antiderivatives of $e^{2x}$, with signs alternating from a plus.
Hint 3/4
The right column is $e^{2x}$, $\frac{e^{2x}}{2}$, $\frac{e^{2x}}{4}$, $\frac{e^{2x}}{8}$, and the diagonal products are $x^{2}\cdot\frac{e^{2x}}{2}$, then $2x\cdot\frac{e^{2x}}{4}$, then $2\cdot\frac{e^{2x}}{8}$.
Hint 4/4
The answer is $\frac{e^{2x}}{4}\left(2x^{2}-2x+1\right)+C$.
Show solution
The table rather than two written-out passes, because the only thing the long form adds here is two more chances to misplace a factor of $\frac{1}{2}$.
Every polynomial-against-exponential answer carries a bracket with alternating signs and shrinking coefficients, so growing coefficients mean a factor of $\frac{1}{a}$ was dropped.
Full exam-style question
Final-style: ∫₁^∞ (ln x)/x² dx, with the limit written outexam format
This is what a complete answer looks like when a question combines the two halves of the section: an integral that has to be set up as a limit, and an antiderivative that only parts can produce.
whether the integral converges, and its exact value if it does
Solution
Parts before the limit, not after. The limit cannot be taken until there is an antiderivative to take it of, and the antiderivative needs one pass of parts with $u=\ln x$, which is the only choice since $\ln x\,dx$ has no elementary $v$.
Declare the integral improper and replace the endpoint
A ceiling without doing the integral: $\ln x\le 2\sqrt{x}$ for $x\ge 1$, so the integrand is at most $2x^{-3/2}$, whose tail integral is $4$, and an answer of $1$ sits under that ceiling.
One pass of parts, one evaluation, one limit. The limit statement at the start and the verdict at the end are part of the answer, not decoration.
Set up the limit, use a technique from this week for the antiderivative, evaluate on the window, and only then let the window grow; an answer that never writes a limit has skipped the part being examined.
Practice
A · concept 3 questions
1§13.1 — when a product does not mean parts●○○○○
A classmate writes a rule on the board before the exam: any integrand that is a product of two functions has to be integrated by parts.
Given
the proposed rule: every product under an integral sign requires integration by parts
Find
True or false, and why?
Hint 1/4
A universal claim is broken by one counterexample, so you are looking for a product that some other method finishes.
Hint 2/4
A substitution finishes $\int f(g(x))g'(x)\,dx$, and that integrand is a product too.
Hint 3/4
Take $\int 2x\cos(x^{2})\,dx$: it is a product, and $u=x^{2}$ finishes it in one line as $\sin(x^{2})+C$.
Hint 4/4
The rule is false; a product only calls for parts when no substitution is available.
Show solution
We look for a counterexample rather than argue in general, because a single explicit integral settles a universal claim and costs one line.
Produce the counterexample
$$u=x^{2},\qquad du=2x\,dx$$
the integrand is a product, but one factor is the derivative of what is inside the other
Differentiate the antiderivative: $\cos(x^{2})\cdot 2x$, the integrand. So the counterexample really is a counterexample, which is the part worth checking before relying on it.
The working rule is the other way round: check for a substitution first, and reach for parts only when that check fails.
2§13.6 — spotting the case with no odd exponent●●○○○
Four trigonometric integrals. Three of them can be started by peeling a single factor; one of them cannot and needs the integrand rewritten before any integration happens.
Given
$\int\sin^{4}x\,dx$
$\int\sin^{3}x\,dx$
$\int\sin^{5}x\cos x\,dx$
$\int\sin x\cos^{3}x\,dx$
Find
Which one needs the power reduction identities?
Hint 1/4
You are looking for the integrand in which neither exponent is odd. Nothing has to be integrated to decide.
Hint 2/4
A peel is available whenever at least one exponent is odd; power reduction is the method for the case where both are even.
Hint 3/4
Write the missing exponent as zero where a function does not appear: $\sin^{4}x$ is $\sin^{4}x\cos^{0}x$, and zero is even.
Hint 4/4
The one that needs power reduction is $\int\sin^{4}x\,dx$.
Show solution
We write in the invisible exponent rather than reasoning about which functions are present, because the parity test needs two numbers and one of them is easy to forget to look for.
Tabulate the parities
$$\sin^{4}x\cos^{0}x:\ (4,0)$$
both even, so no peel exists and the identity is the only route
$$\sin^{3}x\cos^{0}x:\ (3,0)$$
the sine exponent is odd, so one $\sin x$ comes off as the differential
$$\sin^{5}x\cos^{1}x:\ (5,1)$$
both odd and positive, so either factor may be peeled and the smaller exponent is the cheaper choice
$$\sin^{1}x\cos^{3}x:\ (1,3)$$
both odd again, with the roles of the two functions exchanged
Answer $$\int\sin^{4}x\,dx$$
Check
Confirm by trying to peel: taking one $\sin x$ out of $\sin^{4}x$ leaves $\sin^{3}x$, an odd power, which the Pythagorean identity cannot convert into a polynomial in cosine. The obstruction is real, not an oversight.
Fill in the zero exponent every time. It costs nothing and it removes the single most common misreading in this family.
3§13.7 — a positive integrand with a negative answer●●●○○
A student writes $\int_{-1}^{1}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{-1}^{1}=-1-1=-2$ and moves on. The arithmetic in that line is correct.
Given
$$\displaystyle\int_{-1}^{1}\frac{dx}{x^{2}}$$
the student's answer: $-2$
Find
What has gone wrong?
Hint 1/4
Before looking for an algebra slip, ask whether the answer can possibly be right: the integrand is positive everywhere it is defined.
Hint 2/4
The evaluation half of the Fundamental Theorem requires the integrand to be continuous on the whole closed interval $[a,b]$.
Hint 3/4
Here the integrand is $\frac{1}{x^{2}}$ on $[-1,1]$, and at $x=0$ it is not defined at all, let alone continuous.
Hint 4/4
The integral is improper because of the blow-up at $x=0$, and once split correctly both halves diverge.
Show solution
We check the hypotheses of the theorem rather than recheck the arithmetic, because the arithmetic is correct and the answer is still impossible.
Locate the failure
$$\frac{1}{x^{2}}\to\infty\ \text{as}\ x\to 0$$
the point of failure is interior, so nothing in the notation warns you about it
$$\int_{-1}^{1}=\int_{-1}^{0}+\int_{0}^{1}$$
the split is compulsory: each piece must carry the bad point as an endpoint
divergent, and one divergent piece is enough to make the whole integral divergent
Answer $$\text{The integral diverges}$$
Check
A check that needs no computation at all: the integrand is positive on the whole interval, so any correct answer must be positive. The student's $-2$ was refuted before the analysis started.
Before applying the Fundamental Theorem, look for points where the integrand is undefined, and look inside the interval and not only at its ends.
B · computation 5 questions
1§13.5 — an odd sine exponent●●●○○
A trigonometric integral where the sine exponent is odd and the cosine exponent is even, so the route is decided before any work starts.
Given
$$\displaystyle\int\sin^{5}x\cos^{2}x\,dx$$
Find
Evaluate the integral.
Hint 1/4
Look at the two exponents and decide which function has the odd one. That decides everything else.
Hint 2/4
The sine exponent $5$ is odd, so peel one $\sin x$ as the differential and substitute $u=\cos x$, with $du=-\sin x\,dx$.
Hint 3/4
What is left is $\sin^{4}x=\left(1-\cos^{2}x\right)^{2}$, so the integral becomes $-\int\left(1-u^{2}\right)^{2}u^{2}\,du$.
Hint 4/4
The answer is $-\frac{\cos^{3}x}{3}+\frac{2\cos^{5}x}{5}-\frac{\cos^{7}x}{7}+C$.
Show solution
We peel a sine because the sine exponent is the odd one; peeling a cosine would leave $\cos^{1}x$, which the identity cannot turn into a polynomial in sine.
Differentiate: $\cos^{2}x\sin x-2\cos^{4}x\sin x+\cos^{6}x\sin x=\cos^{2}x\sin x\left(1-\cos^{2}x\right)^{2}=\cos^{2}x\sin^{5}x$, the integrand.
A three-term answer with alternating signs is the shape to expect here; a two-term answer would mean the square was never expanded.
2§13.6 — an even power on its own●●●○○
No odd exponent anywhere, so the identity has to be applied twice, and the angle doubles each time.
Given
$$\displaystyle\int\sin^{4}x\,dx$$
Find
Evaluate the integral.
Hint 1/4
There is nothing to peel, so the integrand has to be rewritten before it can be integrated at all.
Hint 2/4
Use $\sin^{2}x=\frac{1-\cos 2x}{2}$ on $\sin^{4}x=\left(\sin^{2}x\right)^{2}$, then expand the square.
Hint 3/4
Expanding gives $\frac{1}{4}\left(1-2\cos 2x+\cos^{2}2x\right)$, and the last term needs the identity again: $\cos^{2}2x=\frac{1+\cos 4x}{2}$.
Hint 4/4
The answer is $\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$.
Show solution
Power reduction rather than the reduction formula: either works, and the formula is quicker if you have it written down, but this route uses only the identity and shows where the three separate terms come from.
Check against the reduction formula, which is a genuinely different route: $\int_{0}^{\pi/2}\sin^{4}x\,dx$ was found earlier to be $\frac{3\pi}{16}$, and this antiderivative gives $\frac{3\pi}{16}-\frac{\sin\pi}{4}+\frac{\sin 2\pi}{32}-0=\frac{3\pi}{16}$. The two agree.
The coefficient of $x$ in the answer is the average value of $\sin^{4}x$, and every even power integrates to its average times $x$, plus waves.
3§13.3 — the returning case with two coefficients●●●●○
Neither factor simplifies under differentiation, so nothing terminates and two passes bring the original integral back.
Given
$$\displaystyle I=\int e^{2x}\cos x\,dx$$
Find
Evaluate the integral by solving for $I$.
Hint 1/4
Do not expect this one to terminate. Name the integral $I$ before you start, because you are going to solve for it.
Hint 2/4
Two passes with the same orientation. Keep the trigonometric factor as $u$ both times, or keep the exponential as $u$ both times, but do not switch.
Hint 3/4
Taking $u=\cos x$ and $dv=e^{2x}dx$ twice gives $I=\frac{e^{2x}\cos x}{2}+\frac{e^{2x}\sin x}{4}-\frac{I}{4}$.
Hint 4/4
Collecting: $\frac{5}{4}I=\frac{e^{2x}}{4}\left(2\cos x+\sin x\right)$, so $I=\frac{e^{2x}}{5}\left(2\cos x+\sin x\right)+C$.
Show solution
We keep the cosine as $u$ on both passes. Switching on the second pass would undo the first exactly and leave the true but useless line $I=I$.
The denominator $5$ is $2^{2}+1^{2}$, and that pattern holds for every integral of this shape, which makes it the fastest available check.
4§13.5 — a tangent standing inside a product of powers●●●○○
A tangent has been left standing next to a power of the sine, so the first move is not a substitution.
Given
$$\displaystyle\int\sin^{4}x\tan x\,dx$$
Find
Evaluate the integral.
Hint 1/4
Decide what the integrand looks like once the tangent is gone, and only then read the exponents.
Hint 2/4
$\tan x=\frac{\sin x}{\cos x}$, so the integrand is $\frac{\sin^{5}x}{\cos x}$. The exponents are $5$ and $-1$, both odd, so this is the peel rather than power reduction, and the positive one is the one peeled.
Hint 3/4
Peel one $\sin x$ off, write $\sin^{4}x=\left(1-\cos^{2}x\right)^{2}$, and set $u=\cos x$ with $du=-\sin x\,dx$. The integral becomes $-\int\frac{\left(1-u^{2}\right)^{2}}{u}\,du$; the integrand you started from was $\sin^{4}x\tan x$.
Hint 4/4
The answer is $-\ln\lvert\cos x\rvert+\cos^{2}x-\frac{\cos^{4}x}{4}+C$.
Show solution
The rewrite comes first and costs one line. After the rewrite both exponents are odd and the peel goes to the positive one. This is the worked example $\int\cos^{2}x\cot x\,dx$ in a mirror: there the cotangent put the sine in the denominator, here the tangent puts the cosine there.
peel the sine and not the cosine: both exponents are odd, and peeling the positive one leaves an even power for the identity to convert, while peeling the cosine out of the denominator leaves $\frac{u^{5}}{1-u^{2}}$ with $u=\sin x$, a rational function waiting for partial fractions
A second route reaches the same place with no substitution at all. Expanding first gives $\sin^{4}x\tan x=\tan x-2\sin x\cos x+\sin x\cos^{3}x$, three standard pieces whose antiderivatives are $-\ln\lvert\cos x\rvert$, $-\sin^{2}x$ and $-\frac{\cos^{4}x}{4}$. That total differs from the boxed answer by $\cos^{2}x+\sin^{2}x=1$, a constant, so the two are the same antiderivative.
The rewrite is the routing step, not a cosmetic one: a tangent hides an odd exponent, because a function in the denominator carries the exponent $-1$ and $-1$ is odd.
5§13.7 — a left-hand tail, and then both ends at once●●●●○
This integrand runs off to the left rather than to the right, and the same antiderivative then has to be measured against the other end too.
Given
$$\displaystyle\int_{-\infty}^{0}xe^{x}\,dx$$
you may use $$\displaystyle\lim_{s\to-\infty}se^{s}=0$$
Find
(a) Write the integral as a limit of ordinary definite integrals and evaluate it.
(b) Decide whether $$\displaystyle\int_{-\infty}^{\infty}xe^{x}\,dx$$ converges, and justify the verdict.
(c) Decide whether $$\displaystyle\int_{-\infty}^{-2}\frac{dx}{\lvert x\rvert^{3}}$$ converges, and if so give its value.
Hint 1/4
In (a) only one endpoint is bad, and it is the lower one. Decide which end moves and which end stays before integrating anything, and remember that $-\infty$ is never substituted into an antiderivative.
Hint 2/4
Replace the bad end by a letter: $\int_{-\infty}^{0}=\lim_{s\to-\infty}\int_{s}^{0}$. The antiderivative is one pass of parts, with $u=x$ and $dv=e^{x}dx$.
Hint 3/4
Parts gives $\int xe^{x}dx=e^{x}(x-1)$, so the window is $\left[e^{x}(x-1)\right]_{s}^{0}=-1-e^{s}(s-1)$. The limit you were given is $\lim_{s\to-\infty}se^{s}=0$, and $\lim_{s\to-\infty}e^{s}=0$ on its own.
Hint 4/4
(a) is $-1$. In (b) the half over $[0,\infty)$ has no finite limit, so the whole integral diverges. In (c) the reflection $u=-x$ gives $\int_{2}^{\infty}u^{-3}\,du$, with $p=3>1$.
Show solution
Parts rather than substitution, because $x$ and $e^{x}$ are unrelated factors and differentiating the $x$ kills it in one pass. The limit is taken after the antiderivative is in hand, never during: $-\infty$ is not a number and cannot be put into a formula.
two bad endpoints, so two separate limits with two separate letters; $0$ is chosen only for convenience and any finite split point gives the same verdict
The reflection $x=-t$ turns the left tail into a right tail and settles part (a) independently: $\int_{-\infty}^{0}xe^{x}dx=-\int_{0}^{\infty}te^{-t}dt$, and that integral is $1$ by one pass of parts, giving $-1$ again. The sign agrees, since $xe^{x}<0$ on all of $(-\infty,0)$, and so does the size: the integrand dips only to $-e^{-1}\approx-0.37$, at $x=-1$.
Convergence at one end says nothing about the other: split at a convenient finite point, run the two limits separately, and stop at the first divergent piece.
C · exam level 5 questions
1§13.3 — the classic returning integral●●●●○
Neither a peel nor a power reduction touches this one, and it is the standard example of an integral that has to be solved for rather than finished.
(a) Split the integrand as $\sec x\cdot\sec^{2}x$ and apply integration by parts once.
(b) Use the identity to turn the resulting integral into something containing $I$ again.
(c) Solve for $I$ and write the antiderivative.
Hint 1/4
Nothing here terminates and nothing substitutes, so expect the integral to come back and plan to solve for it.
Hint 2/4
Split as $u=\sec x$ and $dv=\sec^{2}x\,dx$, because $\sec^{2}x$ is the derivative of $\tan x$ and so $v$ is immediate.
Hint 3/4
Then $du=\sec x\tan x\,dx$ and $v=\tan x$, giving $I=\sec x\tan x-\int\sec x\tan^{2}x\,dx$, and $\tan^{2}x=\sec^{2}x-1$ splits that integral into $I$ and $\int\sec x\,dx$.
We split off $\sec^{2}x$ as $dv$ rather than $\sec x$, because $\sec^{2}x$ is a derivative we recognise, so $v=\tan x$ costs nothing; the reverse split would need $\int\sec x\,dx$ as $v$ and would produce a worse integral.
A numerical check on $\left[0,\frac{\pi}{4}\right]$: the formula gives $\approx 1.1478$, and $\sec^{3}x$ runs from $1$ to $2\sqrt2\approx 2.83$ over an interval of length $0.785$, so that size is right.
The same three moves, parts then identity then solve, handle every odd power of secant.
2§13.4 — a reduction formula used twice●●●○○
A definite integral over the quarter period, where the boundary term of the reduction formula vanishes at both ends. Deriving the formula from scratch is not required; using it correctly is.
First decide what happens to the explicit term of the reduction formula at the two endpoints, because if it vanishes the rest is arithmetic.
Hint 2/4
At $x=\frac{\pi}{2}$ the factor $\cos x$ is zero and at $x=0$ the factor $\sin^{n-1}x$ is zero, so the definite version is $J_{n}=\frac{n-1}{n}J_{n-2}$.
Hint 3/4
Apply it three times starting from $n=6$: $J_{6}=\frac{5}{6}J_{4}$, $J_{4}=\frac{3}{4}J_{2}$, $J_{2}=\frac{1}{2}J_{0}$, and $J_{0}=\frac{\pi}{2}$.
The recursion rather than three power reductions: expanding $\left(\frac{1-\cos 2x}{2}\right)^{3}$ would work and would take a page, with a $\cos^{3}2x$ appearing in the middle that needs a peel of its own.
Two independent checks. First, monotonicity: $\sin^{6}x\le\sin^{4}x$ on this interval, so $J_{6}$ must be below $J_{4}=\frac{3\pi}{16}\approx 0.589$, and $\frac{5\pi}{32}\approx 0.491$ is. Second, size: the integrand is below $1$ on an interval of length $\frac{\pi}{2}\approx 1.571$, so the answer must be below that, and it is.
For even $n$ the answer is the product of the odd numbers over the product of the even numbers, times $\frac{\pi}{2}$.
3§13.7 — the other end of the reference family●●●○○
The tail family was worked out in the notes above. This question is about its mirror image, where the trouble is at zero rather than at infinity, and the inequality points the other way.
Given
$$\displaystyle\int_{0}^{1}\frac{dx}{x^{p}}$$, with $p>0$
Find
For which values of $p$ does this integral converge?
Hint 1/4
Do not reuse the tail condition from memory. The bad point here is $x=0$, and the question is how tall the spike is, not how fat the tail is.
Hint 2/4
Write the integral as $\lim_{t\to 0^{+}}\int_{t}^{1}x^{-p}\,dx$ and evaluate the window with the power rule, keeping $p$ as a letter.
Hint 3/4
For $p\neq 1$ that gives $\frac{1-t^{1-p}}{1-p}$, and the verdict turns on whether $t^{1-p}$ stays finite as $t\to 0^{+}$.
Hint 4/4
It converges exactly when $p<1$, and then its value is $\frac{1}{1-p}$.
Show solution
We keep $p$ as a letter for the same reason as before: one computation settles every case, and testing values would leave the boundary case unresolved anyway.
Test against a case computed independently: $p=\frac{1}{2}$ gives $\frac{1}{1-\frac12}=2$, and the worked example earlier found $\int_{0}^{1}\frac{dx}{\sqrt{x}}=2$ directly. The two agree.
Both halves of the family have the same boundary exponent $1$ and it belongs to neither of them. Far out, you want a big exponent; near a spike, you want a small one.
4§13.7 — parts and a limit, on a steeper power●●●●○
The same shape as the worked exam question, with the denominator raised one power higher. Every step changes by a constant, and the verdict does not change at all.
(a) Write the integral as a limit of ordinary definite integrals.
(b) Find the antiderivative by parts.
(c) Evaluate on the window and take the limit, stating the verdict and the value.
Hint 1/4
The infinite endpoint has to be replaced by a letter before anything is integrated, and the verdict at the end is part of the answer, not an afterthought.
Hint 2/4
For the antiderivative use parts with $u=\ln x$ and $dv=x^{-3}dx$, since a logarithm outranks a power and $\ln x\,dx$ has no elementary antiderivative anyway.
Hint 3/4
Then $du=\frac{dx}{x}$ and $v=-\frac{1}{2x^{2}}$, so $\int\frac{\ln x}{x^{3}}dx=-\frac{\ln x}{2x^{2}}+\frac{1}{2}\int x^{-3}dx=-\frac{\ln x}{2x^{2}}-\frac{1}{4x^{2}}$.
Hint 4/4
Evaluating from $1$ to $t$ gives $\frac{1}{4}-\frac{2\ln t+1}{4t^{2}}$, and both subtracted pieces vanish, so the value is $\frac{1}{4}$.
Show solution
Parts with $u=\ln x$, exactly as in the exam example; the only structural difference is that $v$ now carries $-\frac{1}{2x^{2}}$ instead of $-\frac{1}{x}$, and every constant downstream follows from that.
Compare with the exam-style example, which had $x^{2}$ in the denominator and value $1$: a steeper denominator must collect less area, and $\frac{1}{4}<1$.
A complete answer of this family has three parts: the limit statement, the antiderivative, and the verdict sentence.
5§13.6 — an even power whose angle carries a coefficient●●●○○
A disk-method volume from earlier in the term, and the set-up is one line. The integrand it hands you is an even power of a cosine whose angle is not $x$, and that coefficient is where the whole question is won or lost.
Given
the region under $y=\cos(\pi x)$ from $x=0$ to $x=\frac{1}{2}$, above the $x$-axis
the region is rotated about the $x$-axis
disk method: $$\displaystyle V=\pi\int_{a}^{b}\left[f(x)\right]^{2}dx$$
Find
Find the exact volume of the solid.
Hint 1/4
Write the volume integral down before thinking about technique at all. The question only starts once the integrand is visible, and what appears is an even power with nothing to peel.
Hint 2/4
$\cos^{2}(ax)=\frac{1+\cos 2ax}{2}$, and the angle that survives the reduction is $2ax$, not $2x$.
Hint 3/4
The region runs from $x=0$ to $x=\frac{1}{2}$ under $y=\cos(\pi x)$, rotated about the $x$-axis, so $V=\pi\int_{0}^{1/2}\cos^{2}(\pi x)\,dx$. Here $a=\pi$, the integrand becomes $\frac{1}{2}+\frac{\cos 2\pi x}{2}$, and $\int\cos 2\pi x\,dx=\frac{\sin 2\pi x}{2\pi}$.
Hint 4/4
The integral comes to $\frac{1}{4}$, so the volume is $\frac{\pi}{4}$.
Show solution
Disks rather than shells, because the axis of rotation is the $x$-axis and the curve is already a function of $x$; and power reduction rather than a substitution, because a square with no odd factor gives a peel nothing to work with.
Set the volume up
$$V=\pi\int_{0}^{1/2}\cos^{2}(\pi x)\,dx$$
the radius of the disk at position $x$ is the height of the curve there, so the square is forced on you by the method and not chosen
Reduce the square at the angle it actually has
$$\cos^{2}(\pi x)=\frac{1+\cos 2\pi x}{2}$$
the identity with $a=\pi$, so the doubled angle is $2\pi x$; writing $\cos 2x$ here is the single most expensive slip available in this question
$\sin\pi=0$, so the wave term contributes nothing across this particular window
Answer $$V=\frac{\pi}{4}$$
Check
Average height, with no antiderivative involved: on $\left[0,\frac{1}{2}\right]$ the graph of $\cos^{2}(\pi x)$ runs from $1$ down to $0$ and is symmetric about $x=\frac{1}{4}$, where its value is $\frac{1}{2}$, so its average is $\frac{1}{2}$ and the integral is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. The same check catches the slip: reducing to $\cos 2x$ would give about $0.46$ for the integral, an average height of $0.92$ on a window where the curve starts at $1$ and ends at $0$, which the picture cannot support.
When the angle carries a coefficient, write the identity out with $a$ in it before putting the number in. And keep the average-height check for volumes: it is fast, it is independent of the algebra, and it is what makes a lost $\pi$ visible.
D · interleaved 3 questions
1§13.2 — an area, with the technique unnamed●●●○○
A region in the plane, described geometrically. Nothing in the wording says which technique to use, and deciding that is the point of the question.
Given
the curve $y=x\sin x$
the region between that curve and the $x$-axis, from $x=0$ to $x=\pi$
$\sin x\ge 0$ on $\left[0,\pi\right]$, so the curve does not dip below the axis there
Find
Find the exact area of the region.
Hint 1/4
Turn the geometric description into a definite integral first, and check the sign of the integrand on the interval so you know whether absolute values are needed.
Hint 2/4
Area between a non-negative curve and the axis is $\int_{a}^{b}y\,dx$; here the integrand is $x\sin x$, and it needs integration by parts with $u=x$.
Hint 3/4
With $u=x$, $dv=\sin x\,dx$, $du=dx$ and $v=-\cos x$, the antiderivative is $-x\cos x+\sin x$.
Hint 4/4
Evaluating from $0$ to $\pi$ gives $\pi$, since $\cos\pi=-1$ and both sine terms vanish.
Show solution
We check the sign before integrating, because area is the integral of the curve only where the curve is above the axis; on $\left[0,\pi\right]$ both $x$ and $\sin x$ are non-negative, so no absolute values are needed and the ordinary integral is the area.
Set up the integral
$$A=\int_{0}^{\pi}x\sin x\,dx$$
the integrand is non-negative on this interval, so the integral is the area rather than a signed area
Integrate by parts
$$u=x,\ dv=\sin x\,dx,\ du=dx,\ v=-\cos x$$
algebraic before trigonometric; substitution is unavailable since the derivative of the inside of $\sin x$ is $1$
$\cos\pi=-1$ turns the boundary term positive, and both sine terms vanish at the endpoints
Answer $$A=\pi$$
Check
A size check with no calculus in it: the hump has height about $1.82$ and width $\pi$, so a triangle-like estimate of $2.9$ is close to the exact $\pi\approx 3.14$.
Convert the description into an integral first and classify the integrand second, because choosing the technique before writing the integral is how the wrong one gets chosen.
2§13.2 — a volume that needs a technique from this week●●●●○
A solid of revolution, set up by the shell method from earlier in the term. The set-up is routine; the integral it produces is not, and that is the part this week supplies.
Given
the region under $y=e^{-x}$ from $x=0$ to $x=1$, above the $x$-axis
Set the volume up first and look at the integrand you get before deciding anything about technique.
Hint 2/4
Rotating about the $y$-axis with vertical strips gives shells of radius $x$ and height $y$, so $V=2\pi\int_{0}^{1}xe^{-x}\,dx$.
Hint 3/4
That integral needs parts with $u=x$ and $dv=e^{-x}dx$, so $v=-e^{-x}$ and $\int_{0}^{1}xe^{-x}dx=\left[-xe^{-x}\right]_{0}^{1}+\int_{0}^{1}e^{-x}dx$.
Hint 4/4
The integral works out to $1-\frac{2}{e}$, so the volume is $2\pi\left(1-\frac{2}{e}\right)$.
Show solution
Shells rather than washers, because the axis of rotation is vertical and the region is described by a function of $x$; washers would need the curve inverted, which brings a logarithm into the integrand for no gain.
Set up the shell integral
$$V=2\pi\int_{0}^{1}x\,e^{-x}\,dx$$
radius $x$ measured from the axis of rotation, height $e^{-x}$ read off the curve
both pieces are numbers, so nothing with an $x$ in it can survive
Assemble
$$V=2\pi\left(1-\frac{2}{e}\right)$$
the constant from the shell formula multiplies the whole integral
Answer $$V=2\pi\left(1-\frac{2}{e}\right)$$
Check
A bounding check: rotating the whole unit square about the $y$-axis gives a cylinder of volume $\pi\approx 3.14$, and the answer $1.66$ is a little over half of that.
Interleaving is the point here: the volume formula is from an earlier block and the integral is from this one, and an exam question is free to combine them without saying so.
3§13.2 — substitute first, then integrate by parts●●●●●
Neither technique alone finishes this one. The list would pick a $dv$ that cannot be antidifferentiated, which is the signal that something has to happen before parts is applied at all.
Given
$$\displaystyle\int_{0}^{1}x^{5}e^{x^{3}}\,dx$$
Find
Evaluate the definite integral exactly.
Hint 1/4
Before choosing $u$ and $dv$, ask whether $e^{x^{3}}$ can be antidifferentiated at all. It cannot, so parts on its own is blocked.
Hint 2/4
Substitute to fix that: with $t=x^{3}$ the exponent becomes linear, and $dt=3x^{2}dx$ takes care of two of the five powers of $x$.
Hint 3/4
Then $x^{5}dx=x^{3}\cdot x^{2}dx=t\cdot\frac{dt}{3}$ and the limits become $t=0$ and $t=1$, so the integral is $\frac{1}{3}\int_{0}^{1}te^{t}\,dt$.
Hint 4/4
That last integral is $\left[e^{t}(t-1)\right]_{0}^{1}=0-(-1)=1$, so the answer is $\frac{1}{3}$.
Show solution
Substitution first, parts second. The order is forced: parts needs a $v$, and $e^{x^{3}}$ has no elementary antiderivative, so the exponent must be linearised before the method has anything to work with.
Substitute to linearise the exponent
$$t=x^{3},\qquad dt=3x^{2}\,dx$$
the exponent is the composite piece; its derivative accounts for two of the five powers of $x$
the two copies of $e$ cancel, which is why the answer is so clean
$$\frac{1}{3}\cdot 1=\frac{1}{3}$$
the constant from the substitution
Answer $$\frac{1}{3}$$
Check
A bound without doing the integral: on $\left[0,1\right]$ we have $e^{x^{3}}\le e$, so the value is at most $e\int_{0}^{1}x^{5}dx=\frac{e}{6}\approx 0.45$, and at least $\int_{0}^{1}x^{5}dx=\frac{1}{6}\approx 0.17$ since $e^{x^{3}}\ge 1$. The answer $\frac{1}{3}\approx 0.33$ sits between them.
When the order picks a $dv$ you cannot antidifferentiate, that is information rather than a dead end: it says a substitution has to come first.
Shaped like the real papers 4 questions
1§13.7 — a final in four links: limit, reference, verdict, value●●●●●
Four linked parts, about twenty minutes on a final. The convergence proof in (c) and the value in (d) are marked as two separate jobs, so answering (c) by pointing at the number found in (d) collects nothing for (c).
$e^{-\sqrt{x}}$ is positive and continuous on $[4,\infty)$
Direct Comparison Test: if $0\le f(x)\le g(x)$ for all $x\ge a$ and $\int_{a}^{\infty}g(x)\,dx$ converges, then $\int_{a}^{\infty}f(x)\,dx$ converges
Find
(a) Show that $$\displaystyle\lim_{x\to\infty}x^{2}e^{-\sqrt{x}}=0$$.
(b) Write $$\displaystyle\int_{4}^{\infty}\frac{dx}{x^{2}}$$ as a limit of ordinary definite integrals, evaluate it, and say whether it converges.
(c) Combine (a) and (b) to show that $I$ converges. Do not evaluate $I$ here.
(d) Now use the substitution $y=\sqrt{x}$ to find the exact value of $I$.
Hint 1/4
Two different questions are being asked about the same integral: whether it is finite, and what it equals. They are marked separately, so (c) has to stand on its own without borrowing anything from (d).
Hint 2/4
The Direct Comparison Test needs three things written down: an inequality $0\le f\le g$, the interval on which that inequality holds, and a reference integral $\int g$ already known to converge. A bound that only holds far out is still a bound, provided you say from where.
Hint 3/4
Part (a) says $x^{2}e^{-\sqrt{x}}$ eventually drops below $1$, which after dividing by $x^{2}$ reads $e^{-\sqrt{x}}\le x^{-2}$ for $x\ge X$. For (d), $y=\sqrt{x}$ gives $x=y^{2}$ and $dx=2y\,dy$, and it moves the lower endpoint from $x=4$ to $y=2$.
Hint 4/4
By (b) the reference tail converges, so the tail of $I$ converges by the Direct Comparison Test and the leftover piece on $[4,X]$ is an ordinary integral. In (d), $\int_{2}^{\sqrt{t}}2ye^{-y}\,dy=6e^{-2}-2(\sqrt{t}+1)e^{-\sqrt{t}}\to\frac{6}{e^{2}}$.
Show solution
The comparison is made against $x^{-2}$ rather than against any other member of the reference family because part (a) hands us exactly that inequality and nothing sharper is needed; any $p>1$ would do, and choosing the one the previous part already proved keeps the argument to three lines. In (d) the substitution comes before the parts, because $e^{-\sqrt{x}}$ has no $u$ and $dv$ split that helps while $2ye^{-y}$ has an obvious one.
four applications of L'Hopital's Rule, each on an $\frac{\infty}{\infty}$ form; the numerator loses a degree every time and the denominator does not change
$$\lim_{x\to\infty}x^{2}e^{-\sqrt{x}}=0$$
the exponential beats the fourth power, which is the fact part (c) will need
the power rule on the window, then the limit; the limit is finite, so the integral converges to $\frac{1}{4}$
(c) The comparison, written the way it is marked
$$\exists\,X\ge 4:\quad x^{2}e^{-\sqrt{x}}\le 1\quad\text{for all }x\ge X$$
a limit equal to $0$ means the quantity eventually stays below any fixed positive number, here $1$; the word eventually is why the line carries an $X$ and not the number $4$
on the closed bounded interval $[4,X]$ the integrand is continuous, so that piece is an ordinary definite integral and a finite number; a finite number plus a convergent tail is convergent
the endpoints are carried through the substitution rather than copied, so the lower endpoint becomes $2$; keeping the old $4$ here is the standard way to lose this mark
the second term is $\frac{2(y+1)}{e^{y}}$ with $y=\sqrt{t}\to\infty$, which goes to $0$ for the same reason as in (a); the limit is finite, so the verdict is that $I$ converges to $\frac{6}{e^{2}}$
Check
$\frac{6}{e^{2}}\approx 0.812$. The integrand is $0.135$ at $x=4$, $0.050$ at $x=9$ and $0.018$ at $x=16$, and trapezoids on $[4,9]$, $[9,16]$, $[16,25]$ give $0.814$, and the tail beyond $x=25$ adds $12e^{-5}$, about $0.081$ more, so the estimate is about $0.89$. Since $\frac{d^{2}}{dx^{2}}e^{-\sqrt{x}}=\frac{1}{4}e^{-\sqrt{x}}\left(x^{-3/2}+x^{-1}\right)>0$, the integrand is convex and the trapezoid estimate should sit above the exact value, which it does; with a mesh this coarse it lands about ten percent high.
2§13.7 — three verdicts, no antiderivatives●●●●●
One block, three verdicts, about fifteen minutes. No value is asked and none of the three integrands has an antiderivative worth chasing, so every mark sits in the inequality, the interval it holds on, and the sentence that names the test.
$$\displaystyle\int_{0}^{1}\frac{dx}{x^{p}}$$ converges exactly when $p<1$, and $$\displaystyle\int_{1}^{\infty}\frac{dx}{x^{p}}$$ converges exactly when $p>1$
$\cos x\ge 1-\frac{x^{2}}{2}$ for every real $x$
Find
Decide for each of the three integrals whether it converges or diverges. You are not asked for any values.
Hint 1/4
Before any technique, mark on each integral where it stops being an ordinary definite integral. One of the three fails in two places at once, and no antiderivative in this list is worth hunting for.
Hint 2/4
The Direct Comparison Test runs in both directions: a positive integrand under a convergent one converges, and a positive integrand above a divergent one diverges. When a single integral is bad at two points, split it at an interior point and settle each piece with its own reference, because the whole converges only if every piece does.
Hint 3/4
For (i), $\sin$ never leaves $[-1,1]$, so the numerator sits between $1$ and $3$. For (ii), the given inequality says $1-\cos x\le\frac{x^{2}}{2}$, which turns the integrand into something at least as big as $\frac{2}{x^{2}}$. For (iii), factor $x^{3}-8=(x-2)\left(x^{2}+2x+4\right)$ near $x=2$, and for large $x$ note that $x\ge 3$ forces $8\le\frac{8}{27}x^{3}$.
Hint 4/4
(i) is dominated by $3x^{-2/3}$ with $p=\frac{2}{3}<1$, so it converges. (ii) dominates $2x^{-2}$ with $p=2>1$ at the bad end $0$, so it diverges. (iii) splits at $x=3$ into a piece dominated by $\frac{1}{2\sqrt{3}}(x-2)^{-1/2}$ and a piece dominated by $\sqrt{\frac{27}{19}}\,x^{-3/2}$, both convergent, so it converges.
Show solution
Each item is settled by a bound and never by an antiderivative, because none of the three has one that helps. The reference is chosen to match the behaviour at the bad point only: near a blow-up at $c$ the useful comparison is with a power of $x-c$, and out at infinity it is with a power of $x$, which is why (iii) has to be cut in two before either reference can be used.
inverting reverses the inequality; the integrand is unbounded at $x=0$ because $1-\cos x\to 0$ there, so this is a even though both endpoints are ordinary numbers
the integrand is unbounded at $x=2$ and the interval is unbounded on the right, so the integral is improper at both ends and each piece must be judged on its own
a larger denominator makes a smaller fraction; near the blow-up the integrand behaves like an inverse square root, and $\int_{2}^{3}(x-2)^{-1/2}\,dx$ converges since the exponent $\frac{1}{2}$ is below $1$
taking square roots of the previous line; $\int_{3}^{\infty}x^{-3/2}\,dx$ converges since $\frac{3}{2}>1$
$$\Rightarrow\ \text{both pieces converge by the Direct Comparison Test, so the whole integral converges}$$
the verdict is reported only after every piece has been settled, since one divergent piece would sink the whole
Check
The sizes agree with the verdicts. In (i) the integrand is between $x^{-2/3}$ and $3x^{-2/3}$, so the total is trapped between $3$ and $9$. In (ii) the integrand is about $\frac{2}{x^{2}}$ near zero, and $\int_{a}^{1}\frac{2\,dx}{x^{2}}=\frac{2}{a}-2\to\infty$. In (iii) the two bounding integrals contribute $\frac{1}{\sqrt{3}}\approx 0.58$ and $\sqrt{\frac{27}{19}}\cdot\frac{2}{\sqrt{3}}\approx 1.38$, so the value is finite and below about $2$.
3§13.7 — a comparison that proves nothing●●●●○
Five minutes, and almost all of the marks live in one sentence. This course grades the reason rather than the verdict, so an argument that reaches its conclusion by a step the test does not license scores nothing even when the conclusion sounds plausible.
A student hands in the following as a complete answer. For $x\ge 1$ we have $x+\sqrt{x}\ge x$, so $\frac{1}{x+\sqrt{x}}\le\frac{1}{x}$. The integral is therefore no larger than $\int_{1}^{\infty}\frac{dx}{x}$, so it converges by comparison.
Direct Comparison Test: for $0\le f\le g$ on $[a,\infty)$, convergence of $\int_{a}^{\infty}g$ forces convergence of $\int_{a}^{\infty}f$, and divergence of $\int_{a}^{\infty}f$ forces divergence of $\int_{a}^{\infty}g$
Find
(a) In one sentence, say why this answer earns no marks even though every inequality written in it is true.
(b) Decide whether the integral converges or diverges, and justify your verdict with a comparison the test actually licenses. Name the test and state the interval on which your inequality holds.
Hint 1/4
Read the last line as a claim about what has been proved rather than about what is true. Then ask the second question separately: is the verdict itself even correct?
Hint 2/4
An upper bound is only worth something when the thing above is finite, and a lower bound is only worth something when the thing below is infinite. Bounding above by $\int_{1}^{\infty}\frac{dx}{x}$ places the integral under something that is already infinite, which excludes nothing.
Hint 3/4
For $x\ge 1$ we have $\sqrt{x}\le x$, so the denominator satisfies $x+\sqrt{x}\le 2x$, and inverting a positive inequality reverses it.
Hint 4/4
That gives $\frac{1}{x+\sqrt{x}}\ge\frac{1}{2x}>0$ on $[1,\infty)$, and $\int_{1}^{\infty}\frac{dx}{2x}$ diverges, so the given integral diverges by the Direct Comparison Test. The student's verdict was wrong as well as unjustified.
Show solution
The faulty argument is left standing rather than corrected line by line, because every line of it is true and only the inference at the end is illegitimate. Fixing it means choosing the other direction of the same test, and the constant $2$ in the new bound is the cheapest one that works on the whole interval $[1,\infty)$ rather than only far out.
this is the one sentence the marks are for: an upper bound by a divergent integral leaves both verdicts open, so the argument establishes nothing at all
the minorant diverges, stated as its own sentence so that the conclusion has something to rest on
$$\Rightarrow\ \int_{1}^{\infty}\frac{dx}{x+\sqrt{x}}\ \text{diverges by the Direct Comparison Test}$$
an integrand above a divergent one cannot have a finite integral
Check
This one can be checked exactly, which the exam version would not be. With $u=\sqrt{x}$, $\int\frac{dx}{x+\sqrt{x}}=\int\frac{2u\,du}{u^{2}+u}=2\ln\left(1+\sqrt{x}\right)+C$, and differentiating returns $\frac{1}{\sqrt{x}\left(1+\sqrt{x}\right)}=\frac{1}{x+\sqrt{x}}$. Then $\int_{1}^{t}=2\ln\left(1+\sqrt{t}\right)-2\ln 2\to\infty$, so the integral really does diverge and the student's verdict was false as well.
4§13.7 — a bad endpoint that the substitution sends to infinity●●●●●
Three parts, about fifteen minutes. The substitution moves both endpoints and turns a blow-up at zero into an infinite tail, and a separate mark sits on carrying the ends through the change of variable instead of copying them.
$$\displaystyle\int_{0}^{1}\frac{dx}{x^{p}}$$ converges exactly when $p<1$
Direct Comparison Test, in the form for a bad left endpoint: if $0\le f\le g$ on $(0,b]$ and $\int_{0}^{b}g$ converges, then $\int_{0}^{b}f$ converges
Find
(a) Say in one sentence why this integral is improper, and write it as a limit of ordinary definite integrals.
(b) Show that it converges, without computing its value.
(c) Evaluate $I$ exactly using the substitution $y=-\ln x$.
Hint 1/4
Check both ends before writing anything. Only one of them causes trouble, and the trouble there is not a division by zero.
Hint 2/4
A comparison near a bad left endpoint is made with $x^{-p}$, and the convergent side of that family at zero is $p<1$. When a substitution is applied to a window with a letter in one endpoint, the letter is pushed through the substitution formula like every other endpoint, and reversing the order of the two ends is what the minus sign from $dx$ pays for.
Hint 3/4
Since $\sqrt{x}\left(\ln x\right)^{2}\to 0$ as $x\to 0^{+}$, there is a $\delta$ with $\left(\ln x\right)^{2}\le\frac{1}{\sqrt{x}}$ on $(0,\delta)$. For the value, $y=-\ln x$ means $x=e^{-y}$ and $dx=-e^{-y}\,dy$, and it sends $x=a$ to $y=-\ln a$ and $x=1$ to $y=0$.
Hint 4/4
The comparison with $p=\frac{1}{2}$ settles (b). In (c) the integral becomes $\int_{0}^{A}y^{2}e^{-y}\,dy=2-\left(A^{2}+2A+2\right)e^{-A}$ with $A=-\ln a\to\infty$, so $I=2$.
Show solution
The verdict is settled before the value and by a different tool, because the comparison costs three lines while the exact evaluation costs eight, and on this course the two are marked separately. The substitution $y=-\ln x$ rather than $y=\ln x$ is chosen so that the new variable runs from $0$ upward through positive values, which keeps the exponential in the familiar decaying form $e^{-y}$.
$\ln x\to-\infty$ at the left end and squaring keeps it unbounded, so the integrand is unbounded on $(0,1]$ even though both endpoints are ordinary numbers; at $x=1$ the integrand is simply $0$, so the right end is fine
a limit equal to $0$ puts the product below $1$ eventually, and dividing by $\sqrt{x}>0$ gives the bound; as in every comparison here, the interval on which it holds is part of the statement
the reference family at zero, on its convergent side; stating this convergence explicitly is a scoring step
$$\Rightarrow\ \int_{0}^{\delta}\left(\ln x\right)^{2}dx\ \text{converges by the Direct Comparison Test, and}\ \int_{\delta}^{1}\left(\ln x\right)^{2}dx\ \text{is an ordinary integral}$$
on $[\delta,1]$ the integrand is continuous, so that piece is a finite number; a convergent piece plus a finite number is convergent, so $I$ converges
(c) The substitution, watching both ends
$$y=-\ln x,\quad x=e^{-y},\quad dx=-e^{-y}\,dy$$
the substitution written in the direction it will be used, so that $dx$ carries the minus sign openly
the endpoints are pushed through the substitution and come out in the reverse order, with the bad end $a\to 0^{+}$ turning into $A\to\infty$; copying $0$ and $1$ across is exactly the error this part is testing
the second pass gives $\int ye^{-y}\,dy=-(y+1)e^{-y}$, and collecting the terms produces the closed form; the table form of parts gives the same three terms in one line
the exponential beats the quadratic, so the bracket vanishes; the limit is finite, so the verdict is that $I$ converges to $2$
Check
Differentiating backwards checks the answer by a second route: since $\frac{d}{dx}\left[x\left(\ln x\right)^{2}-2x\ln x+2x\right]=\left(\ln x\right)^{2}$, the window integral equals $\left[x\left(\ln x\right)^{2}-2x\ln x+2x\right]_{a}^{1}=2-a\left(\ln a\right)^{2}+2a\ln a-2a\to 2$. Numerically the integrand exceeds $1$ only on $\left(0,e^{-1}\right)$, where it grows without bound slowly, so an average height of $2$ over $(0,1)$ is the right size.
Mistake ledger (23 entries)
⚠ Antidifferentiating the two factors separately
the sum rule does not extend to products
wrong$$\int x\cos x\,dx=\frac{x^{2}}{2}\sin x+C$$
right$$\int x\cos x\,dx=x\sin x+\cos x+C$$
⚠ Letting the minus sign cover only the first term
the minus belongs to the whole traded integral
wrong$$\int x\cos x\,dx=x\sin x+\int\sin x\,dx$$
right$$\int x\cos x\,dx=x\sin x-\int\sin x\,dx$$
⚠ Picking a dv you cannot antidifferentiate
the order was applied before checking that $v$ exists
right$$\text{every piece must converge, or the whole diverges}$$
⚠ Reducing a scaled angle as though the angle were x
the reduced angle is $2ax$, where $a$ is whatever sits inside the function, so a coefficient in the angle survives the identity and also produces the $\frac{1}{4a}$ in the antiderivative
wrong$$\cos^{2}(\pi x)=\frac{1+\cos 2x}{2}$$
right$$\cos^{2}(\pi x)=\frac{1+\cos 2\pi x}{2}$$
⚠ Chasing both infinite ends with a single letter
the definition splits the interval and takes two independent limits, and both have to be finite; a symmetric window cancels the two halves against each other and reports a value that the integral does not have
Both exponents even, or two different angles multiplied. The same-angle grouping $\sin x\cos x=\frac{\sin 2x}{2}$ is the exception on this card: its two exponents are odd, and it earns its place only because it shortens $\sin^{2}x\cos^{2}x$; standing alone it is faster to peel. Every application doubles the angle, and with two different angles present each term is reduced separately.
Both functions continuous and $0\le f\le g$ on an interval reaching the bad point. The inequality only has to hold near that point, because what is left over is an ordinary definite integral of a continuous function. Use it whenever the question asks for a verdict and not a value. An upper bound settles nothing unless the majorant converges, and a lower bound settles nothing unless the minorant diverges. A complete answer writes three separate things: the name of the test, the inequality together with the interval it holds on, and the verdict of the reference integral as its own sentence with the integral sign and both endpoints on it.
Limit Comparison Test
$$L=\lim\frac{f(x)}{g(x)}\ \text{at the bad point},\quad f,g>0:\qquad\begin{cases}0<L<\infty & \text{both integrals share one verdict}\\L=0 & \int g\ \text{converges}\ \Longrightarrow\ \int f\ \text{converges}\\L=\infty & \int g\ \text{diverges}\ \Longrightarrow\ \int f\ \text{diverges}\end{cases}$$
$f>0$ and $g>0$ near the bad point, and the limit is taken at that point, whether it is $\infty$ or a finite endpoint. Use it when the leading behaviour is obvious but an honest inequality is awkward to write, since the middle case needs no inequality at all. It is not lectured in this course, but the 2023 final's key states the $L=0$ half in full and opens a parallel full-credit mark scheme for it beside the direct test, so it is a legitimate route as long as the limit and the reference integral are both written out.
The p test with the singularity moved off the origin
$a<b$ both finite in the first two, $a>0$ in the third. The exponent is read off the factor that actually vanishes, so $(x-a)$ and not $x$: a blow-up at $x=3$ is compared with $(x-3)^{-p}$, and the dividing line stays at $p=1$. Use it as the reference integral in every comparison at a finite bad endpoint, which is where most of the pool's convergence questions have their singularity.
On any interval avoiding the multiples of $\pi$; the two written forms of the cosecant answer differ by a constant, so either is accepted. These belong beside $\int\sec x\,dx$ and $\int\tan x\,dx$ and are expected to be memorised to the same standard: a secant substitution whose numerator is a bare constant lands on $\int\csc\alpha\,d\alpha$, and the pool's key writes the answer with no working shown for it.
Each holds for every value in its stated range, with equality at a single point. Use them at the moment a comparison is needed and the integrand is not a power: they are what replaces a logarithm, an exponential or an oscillating factor by a power of $x$ or of $x-a$, and the power family is the only one whose convergence you know outright. Inverting a positive inequality reverses it, so a bound on the denominator becomes the opposite bound on the fraction.
Addition formulas, and the double angle read as a difference of squares
Valid for all $A$, $B$ and $x$. Use the addition formulas to open a shifted angle such as $x+\frac{\pi}{4}$ before any peeling or power reduction is attempted, and the factored double angle when a $\cos 2x$ standing in a numerator has to be recognised as the derivative of a $\sin x+\cos x$ standing in a denominator. This is the case the peel and the product-to-sum lines both miss, because it carries a power and two different angles at once.
The other four trigonometric functions, and two answers that look different
The first pair needs $\cos x\neq 0$, the second pair $\sin x\neq 0$. Use the definitions at the start, before a route is chosen, because none of the sine and cosine branches in this block mentions a tangent and an unrewritten integrand matches none of them. The tangent and secant family is the exception: it keeps its own box. Use the two logarithm identities at the end, when your answer and a printed one differ only by which reciprocal was named.
$a\neq 0$, angle in radians. Read $a$ off whatever sits inside the function, so $a=\pi$ in $\cos^{2}(\pi x)$ and $a=\frac{1}{2}$ in $\sin^{2}\frac{x}{2}$; the reduced angle is then $2ax$, which for a half angle is $x$ itself. The $\frac{1}{4a}$ is the only difference from the $a=1$ version, and it is the piece that goes missing when the memorised formula is copied without looking at the inner angle.
Improper at both ends, the third line of Type I, and the left-hand tail
As long as the integrand is continuous between the two ends, any finite $c$ may be used and the verdict does not depend on which one; if it also blows up inside, split there too, so that every piece carries exactly one bad point. The pieces are separate integrals with separate limits and all must converge; a convergent half never rescues a divergent one. A single symmetric window $\lim_{t\to\infty}\int_{-t}^{t}$ is not the definition and hands back a value for divergent integrals, $\int_{-\infty}^{\infty}x\,dx$ being the standard example. The left tail is the right tail reflected, since $u=-x$ turns $\int_{-\infty}^{-1}$ into $\int_{1}^{\infty}$, so the dividing line stays at $p=1$ once the exponent is read off $\lvert x\rvert$.
Check yourself
Close the page and write down from memory: the parts formula with its minus sign in the right place, the five classes of the priority order, the two things that can happen when parts is applied twice, the question you ask about a product of a sine and a cosine power, the two identities for the case where that question is answered no, and the two conditions on $p$ with the end of the interval each one belongs to.
Say in one sentence why $\int 2x\cos(x^{2})\,dx$ and $\int x\cos x\,dx$ take different routes.
c-parts
Name $u$ and $dv$ for $\int x^{3}\arctan x\,dx$ without integrating, and say what goes wrong with the other choice.
c-choosing-u
Write the first three rows of the table for $\int x^{2}\sin 3x\,dx$, including the fractions in the right-hand column.
c-parts-repeat
Explain why the explicit term of the sine reduction formula vanishes on $\left[0,\frac{\pi}{2}\right]$ but not on $\left[0,\frac{\pi}{3}\right]$.
c-parts-definite
State which factor you would peel in $\int\sin^{7}x\cos^{5}x\,dx$ and why that choice is cheaper than the other one.
c-trig-odd
Write $\int\cos^{4}x\,dx$ down to the point where every term is ready to integrate, and say where the angle $4x$ came from.
c-trig-even
Give the verdict for $\int_{1}^{\infty}x^{-0.9}\,dx$ and for $\int_{0}^{1}x^{-0.9}\,dx$, and say why they differ.
c-improper
Glossary (16 terms)
integration by partskısmi integrasyon
The identity $\int u\,dv=uv-\int v\,du$, obtained by integrating the product rule. It exchanges one integral for another rather than evaluating anything.
LIATE
The priority order logarithmic, inverse trigonometric, algebraic, trigonometric, exponential, used to decide which factor of an integrand becomes $u$. A habit with a reason behind it, not a theorem.
tablo yöntemi
A two-column layout for repeated integration by parts, used when one factor is a polynomial whose derivatives reach zero. The answer is the sum of the diagonal products with alternating signs.
boundary termsınır terimi
The piece $\bigl[uv\bigr]_{a}^{b}$ of a definite integration by parts. It is a number as soon as it is written, and it is never integrated again.
returning integral
An integral that reappears on the right-hand side after two passes of parts, so that the line becomes an equation to be solved for it rather than an integration to be continued.
reduction formulaindirgeme formülü
A formula that expresses an integral in terms of the same integral with a smaller exponent, applied repeatedly until a basic form is left.
power reduction identity
One of $\sin^{2}x=\frac{1-\cos 2x}{2}$ and $\cos^{2}x=\frac{1+\cos 2x}{2}$, which replace a square by a constant plus a wave at twice the angle.
One of the three identities that rewrite a product of sines and cosines of different angles as a sum, so that the pieces can be integrated one at a time.
Pythagorean identityPisagor özdeşliği
The relation $\sin^{2}x+\cos^{2}x=1$, together with $1+\tan^{2}x=\sec^{2}x$, used to convert an even power of one trigonometric function into a polynomial in another.
peeling a factor
Setting one $\sin x$ or one $\cos x$ aside next to the $dx$ so that it can serve as the differential of a substitution, which is possible exactly when that function carries an odd exponent.
improper integral
A definite integral with an infinite endpoint or an unbounded integrand, defined as the limit of ordinary definite integrals.
Type I improper integral
One that is improper because the interval is unbounded, so that at least one limit of integration is infinite.
Type II improper integral
One that is improper because the integrand is unbounded somewhere on the interval, whether at an endpoint or strictly inside it.
convergentyakınsak
Said of an improper integral whose defining limit exists and is finite. The value reported is that limit.
divergentıraksak
Said of an improper integral whose defining limit is infinite or fails to exist. No value is reported, only the verdict.
The two anchor families $\int_{1}^{\infty}x^{-p}\,dx$, convergent exactly when $p>1$, and $\int_{0}^{1}x^{-p}\,dx$, convergent exactly when $p<1$.
What comes next
§14 · Trig substitution, partial fractions, numerical and improper integrals
The next block finishes the toolbox: a substitution that removes a square root, a decomposition that breaks a ratio of polynomials into pieces you already know, and the rules that approximate an integral when no antiderivative exists. The improper integrals started here come back there as well, because a partial fraction decomposition is often the missing antiderivative inside the limit.
Sources
James Stewart, Calculus, Ninth Edition — sections 7.1, 7.2 and 7.8 The statements of the parts formula, the trigonometric cases and the definition of an improper integral are the ones this book gives; every worked function on this page is a different one.
Course syllabus, week 13: Techniques of Integration 7.1, 7.8, 7.2 The three section numbers used on this page are the three the syllabus line contains, and the assessment weights quoted on the card come from the same syllabus.
Every numerical value on this page Each antiderivative was differentiated back and each definite value recomputed independently before publication, including the window table for the reference family.