14Trig substitution, partial fractions, numerical and improper integrals

Write the expression under the root as $a^{2}-x^{2}$, $a^{2}+x^{2}$ or $x^{2}-a^{2}$, completing the square first if there is a linear term. $a$ is the square root of the constant.

Replace $x$, replace $dx$, and simplify the root by the matching identity. Leaving $dx$ behind is the most common single error.

What is left is a trigonometric integral of the kind handled in the previous section: peel a factor, use an identity, or recognise a standard form such as $\int\sec\theta\,d\theta$.

The substitution itself gives the triangle: $x=a\sin\theta$ means opposite $x$, hypotenuse $a$. Read every trigonometric function of $\theta$ off it.

Rewrite the answer in $x$, add $+C$, and differentiate it once to confirm you get the integrand back.

$$\displaystyle\int\frac{A}{x-a}\,dx=A\ln\lvert x-a\rvert+C$$. Bars, always: the argument changes sign across $a$.

$$\displaystyle\int\frac{A}{(x-a)^{k}}\,dx=\frac{-A}{(k-1)(x-a)^{k-1}}+C$$ for $k\ge2$. A power, not a logarithm.

$$\displaystyle\frac{Bx+C}{x^{2}+a^{2}}$$ splits into the part that is a multiple of the derivative $2x$, giving $\frac{B}{2}\ln(x^{2}+a^{2})$, and the constant left over, giving $\frac{C}{a}\arctan\frac{x}{a}$.

Complete the square first, shift with $u=x+\frac{b}{2}$, then use step 3 on the shifted expression.

Say out loud which endpoint is bad and why. If a blow-up sits strictly inside the interval, split there first; both halves have to converge.

Replace the bad endpoint by $t$ and integrate with the techniques of this section. If the separate limits would give $\infty-\infty$, as two logarithms of unbounded arguments do, combine before taking the limit. That move is a repair, not a routine: when each piece already has a finite limit, as a pair of arctangents does, take the limits separately and combine nothing.

Find $g$ with $0\le f\le g$ and a known verdict for $\int g$; a convergent $g$ drags $f$ with it. In the other direction, a divergent minorant forces divergence. The comparator is usually a $p$ integral.

Either "converges to $L$" with the number, or "diverges". A comparison gives the first word only, never the number.

Infinite endpoints and points where the integrand blows up both count, and one integral can carry both at once. Before comparing, check that the point is genuinely bad: if the integrand stays bounded near it, as $\frac{\sin x}{x}$ does at $0$, the integral is not improper there at all, and the honest answer is that it converges because it is an ordinary integral of a bounded function.

Break the interval at convenient interior points so that every piece owns exactly one bad point. The whole integral converges only if every piece does, and a single divergent piece settles the question. A short piece next to a singularity and an infinite tail almost never accept the same comparator, so expect to build two.

On a tail, keep the highest power in each factor and discard the rest: $\frac{x}{\sqrt[3]{x^{7}-9x^{5}}}$ behaves like $x^{-4/3}$. Near a finite bad point $c$, keep the factor that vanishes: the same integrand behaves like a constant times $(x-c)^{-1/3}$. The guess names a $p$, and the matching $p$ test names the verdict you are trying to prove. In symbols, so that the exponent never has to be re-estimated: with $\deg P=n$ and leading coefficient $a_{n}$, $\lvert P\rvert^{-1/m}$ behaves like $\lvert a_{n}\rvert^{-1/m}\lvert x\rvert^{-n/m}$ in a tail, and like a constant times $\lvert x-c\rvert^{-k/m}$ at a root $c$ of multiplicity $k$. The tail converges when $\frac{n}{m}>1$ and the root when $\frac{k}{m}<1$.

The test needs $0\le f\le g$ or $0\le g\le f$ on the piece, never an approximation. Either quote an elementary bound that holds throughout, such as $\ln x\le x-1$ or $x+3\le2x$ for $x\ge3$, or convert a limit into an inequality on a smaller piece and say so in writing: from $\frac{f}{g}\to1$ there is an $X$ with $f\ge\frac{g}{2}$ for $x>X$, and at a finite endpoint there is a $\delta$ with the inequality holding for $c<x<c+\delta$. Cutting the piece down like this costs nothing, because the removed part is an ordinary integral of a continuous function.

When no clean inequality appears, compute $\lim\frac{f}{g}$, say which case of the test applies, and quote the conclusion. Write the statement of the test out before using it; it is not in the textbook, and the key in the pool awards its marks only to an answer that names it.

Four things are marked one by one, and they are marked separately even when the verdict itself is right.

1. The name of the test, written out: Direct Comparison Test, or Limit Comparison Test.
2. The comparator's own verdict, as a clean sentence of its own with the integral sign and both endpoints on it, such as “$\int_{2}^{\infty}\frac{dx}{x(\ln x)^{2}}$ converges”.
3. The inequality, together with the range it holds on: “for $x\ge X$”, or “for $c<x<c+\delta$”.
4. A closing sentence that ties the three together and states the verdict for the integral you were actually asked about.

A correct verdict with these merged into one scribble scores a fraction of the marks, and pointing at a later part of the question where the integral is evaluated exactly scores nothing when the question said to combine the previous parts.

If the integrand is a rational expression in $x^{1/n_{1}},\dots,x^{1/n_{k}}$, put $u=x^{1/n}$ with $n=\operatorname{lcm}(n_{1},\dots,n_{k})$, so that $x=u^{n}$ and $dx=n\,u^{n-1}\,du$. Every radical becomes a whole power at once. For $\sqrt{x}$ and $\sqrt[3]{x}$ the exponent is $6$; a smaller $n$ leaves one of the two radicals fractional, and a larger one only inflates the degrees.

For a single $\sqrt{g(x)}$ put $u=\sqrt{g(x)}$, so $u^{2}=g(x)$ and $2u\,du=g'(x)\,dx$. Differentiating the squared relation is cheaper than differentiating the root and it hands over the piece of $dx$ you need. This is the second route the papers demand next to the trigonometric one, and the two answers then have to be reconciled.

Put $t=\tan\frac{x}{2}$ on an interval where that is defined. Then $\sin x=\frac{2t}{1+t^{2}}$, $\cos x=\frac{1-t^{2}}{1+t^{2}}$ and $dx=\frac{2\,dt}{1+t^{2}}$, and any ratio built from sines and cosines becomes a ratio of polynomials in $t$. The $dx$ formula is the one that gets forgotten.

Cancel what the substitution made cancellable, check whether the numerator degree is still at least the denominator degree, divide if it is, and only then write a partial fraction template on the remainder. Almost every one of these substitutions produces an improper fraction, so the division is the normal case rather than the exception.

Each of these substitutions is a bijection only on a stated interval, usually $x>0$ for $u=x^{1/n}$; that is what makes the return trip legitimate and what lets absolute value bars be dropped. Rewrite every power of the new variable in terms of $x$ and finish with $+C$.

An odd index radical is defined for negative arguments and carries their sign: $\sqrt[3]{-8}=-2$, so $P(x)^{-1/3}$ is a perfectly ordinary real number wherever $P(x)\neq0$, whatever the sign of $P$. An even index radical is not: $\sqrt{P}$ needs $P\ge0$ and the interval has to be cut where that fails. So a cube root question is a question about signs and never about the domain, and a square root question is the other way round.

Write $P(x)=(x-c)^{k}Q(x)$ with $Q(c)\neq0$. The sign flips at $c$ exactly when $k$ is odd; an even multiplicity touches the axis and comes straight back, and no split is needed there. Far out the sign is that of the leading coefficient as $x\to+\infty$, and that of $a_{n}(-1)^{n}$ as $x\to-\infty$. A polynomial of degree $n$ has at most $n$ real roots, so the sign changes are finite in number and every one of them can be listed.

On each piece between consecutive sign changes $f$ keeps one sign throughout, so either $f$ or $-f$ is non negative there and the direct comparison test applies to it unchanged. The whole integral converges only if every piece does. This is what the sign warning was pointing at: finitely many sign changes are not a dead end, they are a splitting instruction. A tail that crosses zero infinitely often, as $\frac{\sin x}{x}$ does, is the genuine dead end, because there is no last piece to stop at.

If $\int\lvert f\rvert$ converges then $\int f$ converges, and $\lvert f\rvert\ge0$ everywhere, so the ordinary comparison machinery runs on it with no splitting at all. This is the shorter road whenever the pieces would need the same comparator anyway. It runs in one direction only: a divergent $\int\lvert f\rvert$ decides nothing about $\int f$, which is exactly why $\int_{1}^{\infty}\frac{\sin x}{x}\,dx$ has to be settled by parts instead of by comparison.

Near a root $c$ of multiplicity $k$, $\lvert P\rvert^{-1/m}$ behaves like a positive constant times $\lvert x-c\rvert^{-k/m}$; in a tail it behaves like $\lvert a_{n}\rvert^{-1/m}\lvert x\rvert^{-n/m}$. Written without the bars the formula quietly assumes $P>0$ and hands a negative quantity to a $p$ test, which is how a negative leading coefficient or an odd degree breaks an otherwise correct estimate. With the bars the conditions are the familiar pair: $\frac{k}{m}<1$ at the root, $\frac{n}{m}>1$ in the tail.

Take $\int_{0}^{3}\left(x^{2}-1\right)^{-1/3}dx$. The integrand is real and negative on $[0,1)$, real and positive on $(1,3]$, and the only bad point is the simple root $x=1$, so $k=1$ and $m=3$. Split at $1$. On $[0,1)$ we have $(x+1)^{-1/3}\le1$, so $\lvert f\rvert\le(1-x)^{-1/3}$ and $\int_{0}^{1}(1-x)^{-1/3}dx=\frac32$. On $(1,3]$ the same bound gives $\int_{1}^{3}(x-1)^{-1/3}dx=\frac32\cdot2^{2/3}$. Both are finite because $p=\frac13<1$, so the integral converges, and $\frac{k}{m}<1$ was the whole story.

An improper integral imposes one requirement for each place it is improper, and the requirements are independent. An infinite endpoint, a finite blow up, and the endpoints of the interval you were given are all separate demands, and an object that clears one of them and fails another is worth nothing. Write them down as a list before writing any candidate.

For a radical of a polynomial the numbers are exponents: convergence in a tail needs $\frac{n}{m}>1$, and convergence at a root of multiplicity $k$ needs $\frac{k}{m}<1$. Each demand becomes a bound on the degree or on a multiplicity. The smallest admissible object is then read off the strongest of those bounds, not off the first one you wrote.

An even index radical demands that its argument stay non negative on the whole interval, and a logarithm demands that its argument stay positive. These are constraints of exactly the same standing as the convergence ones and they are the ones most often forgotten, because they come from the symbols rather than from the question. Sometimes they are what makes the answer impossible: if the collected inequalities contradict each other, the honest answer is that no such object exists, and it needs the same two lines of reasoning as an example would.

Take the bare power the tail asks for and then remove the finite bad points rather than trying to satisfy them: adding a positive constant to a polynomial usually deletes every root at once and costs no degree. Do not decorate, and do not leave a free parameter standing, because a family is not an example until the parameter has been fixed to a number.

Run the ordinary analysis on the object you produced, as if somebody else had handed it to you, and write the verdict out. Then show that the next cheaper candidate fails, with a genuine inequality and a divergent comparator. A smallest degree question pays for both halves, and an example on its own answers only the easier one.

A numerical check catches a dropped constant factor, a wrong $a$, a sign, and a missing term. It cannot catch a missing $+C$, and it will not distinguish two answers that differ by less than the accuracy you bought. Decide which of those you are worried about before choosing how much work to spend.

On an interval where the integrand only decreases, the left sum is above the true value and the right sum is below it, so the pair straddles the answer with no error formula needed at all. Four strips is usually enough to catch a factor of two. Where the integrand is not monotone, use Simpson's rule and its error bound instead, which is the same idea with the interval width doing the bracketing.

Cut the integral at a convenient point, sample the finite part and dominate the tail by a $p$ integral you can evaluate exactly. The tail bound goes into the bracket as a width, not as a value: the true tail is somewhere between $0$ and the bound, and both ends of the bracket move accordingly.

Write the bracket down as two numbers. A closed form outside it is wrong and the algebra has to be reopened; a closed form inside it has survived the check and nothing more, since a coarse bracket rules out gross errors and not small ones. Reporting agreement to more digits than the bracket carries is the way this check is usually oversold.

For $\int_{0}^{\infty}\frac{dx}{(x^{2}+4)(2x^{2}+1)}$ the bracket built from four strips on $[0,2]$ and a tail bound of $\frac{1}{48}$ runs from $0.133$ to $0.273$. The closed form $\frac{\pi\left(2\sqrt2-1\right)}{28}\approx0.205$ sits inside it. The answer $\frac{3\pi}{28}\approx0.337$, which is what comes out of reading $a=1$ off $2x^{2}+1$, sits outside it and is rejected without any algebra being re-read.