Integration Techniques

2.1 Trigonometric Integrals

Sine and cosine

These integrals are of the form

$$\int {\cos }^{m}x{\sin }^{n}xdx,$$

where $m$ and $n$ are non-negative integers. There are two cases:
(i). either $m$ or $n$ (or both) are odd;
(ii). both $m$ and $n$ are even.

For (i). suppose that $m$ is odd. Then, we can use the substitution $u=\sin x$ with the identity

$${\sin }^{2}x+{\cos }^{2}x=1$$

to evaluate the integral.
Example. Evaluate the integral $\int {\cos }^{3}x{\sin }^{4}xdx$
We use the substitution $u=\sin x,du=\cos xdx$ to yield

$$\begin{array}{rl}\int {\cos }^{3}x{\sin }^{4}xdx& =\int {\cos }^{2}x{\sin }^{4}x\cos xdx\\ & =\int (1-{\sin }^{2}x){\sin }^{4}x\cos xdx\\ & =\int (1-u^2)u^{4}du\\ & =\int u^4-u^{6}du\\ & =\frac{u^5}{5}-\frac{u^7}{7}+C\\ & =\frac{{\sin }^{5}x}{5}-\frac{{\sin }^{7}x}{7}+C\end{array}$$

If $n$ is odd then we use the substitution $u=\cos x$. If both $n$ and $m$ are odd, then either substitution will work.

Example. Evaluate the integral $\int {\cos }^{6}x{\sin }^{5}xdx$.
This time we use $u=\cos x,du=-\sin xdx$.

$$\begin{array}{rl}\int {\cos }^{6}x{\sin }^{5}xdx& =-\int {\cos }^{6}x{\sin }^{4}x(-\sin x)dx\\ & =-\int {\cos }^{6}x(1-{\cos }^{2}x{)}^2(-\sin x)dx\\ & =-\int u^6(1-u^2{)}^{2}du\\ & =-\int u^6-2u^8+u^{10}du\\ & =-(\frac{u^7}{7}-\frac{2u^9}{9}+\frac{u^{11}}{11})+C\\ & =-\frac{{\cos }^{7}x}{7}+\frac{2{\cos }^{9}x}{9}-\frac{{\cos }^{11}x}{11}+C\end{array}$$

For (ii). where both $m$ and $n$ are even we use an entirely different approach. This time we use the identities

$$\begin{array}{rlrlr}{\cos }^{2}x=\frac{1+\cos 2x}{2}& & \text{ and }& & {\sin }^{2}x=\frac{1-\cos 2x}{2}\end{array}$$

to change the integrals into the form

$$\int {\cos }^{k}2xdx.$$

We can then repeat the methods of either case until the integral in the sum is easy to compute.

Example. Evaluate $\int {\sin }^{2}x{\cos }^{4}xdx$.
Using the identities,

$$\begin{array}{rl}\int {\sin }^{2}x{\cos }^{4}xdx& =\int \left(\frac{1-\cos 2x}{2}\right){\left(\frac{1+\cos 2x}{2}\right)}^{2}dx\\ & =\frac{1}{8}\int (1-\cos 2x)(1+\cos 2x{)}^{2}dx\\ & =\frac{1}{8}\int 1+\cos 2x-{\cos }^{2}2x-{\cos }^{3}2xdx\\ & =\frac{x}{8}+\frac{\sin 2x}{16}-\frac{1}{8}\int {\cos }^{2}2xdx-\frac{1}{8}\int {\cos }^{3}2xdx.\end{array}$$

The first integrand of is an even power of $\cos 2x$ and is evaluated using the first identity:

$$\begin{array}{rl}\int {\cos }^{2}2xdx& =\frac{1}{2}\int 1+\cos 4xdx\\ & =\frac{x}{2}+\frac{\sin 4x}{8}+C_1.\end{array}$$

The second integrand of (2.3) is an odd power of $\cos 2x$. The substitution

$$u=\sin 2x,du=2\cos 2xdx$$

gives

$$\begin{array}{rl}\int {\cos }^{3}2xdx& =\frac{1}{2}\int (1-{\sin }^{2}2x)2\cos 2xdx\\ & =\frac{1}{2}\int 1-u^{2}du\\ & =\frac{u}{2}-\frac{u^3}{6}+C_2\\ & =\frac{\sin 2x}{2}-\frac{{\sin }^{3}2x}{6}+C_2.\end{array}$$

Thus we obtain

$$\begin{array}{rl}\int {\sin }^{2}x{\cos }^{4}xdx& =\frac{x}{8}+\frac{\sin 2x}{16}-\frac{x}{16}-\frac{\sin 4x}{64}-\frac{\sin 2x}{16}+\frac{{\sin }^{3}2x}{48}+C\\ & =\frac{x}{16}-\frac{\sin 4x}{64}+\frac{{\sin }^{3}2x}{48}+C.\end{array}$$

Integrating multiple angles of sine and cosine

These integrals consist of integrals in the form:

$$\begin{array}{rlrlrlr}\int \cos mx\sin nxdx,& & \int \cos mx\cos nx& & \text{ or }& & \int \sin mx\sin nx,\end{array}$$

where $m$ and $n$ are real numbers.

$$\begin{array}{rl}\sin A\cos B& =\frac{1}{2}(\sin (A+B)+\sin (A-B))\\ \cos A\cos B& =\frac{1}{2}(\cos (A-B)+\cos (A+B))\\ \sin A\sin B& =\frac{1}{2}(\cos (A-B)-\cos (A+B))\end{array}$$

Example. Evaluate $\int \cos 5x\cos 3xdx.$
Using the identity,

$$\begin{array}{rl}\int \cos 5x\cos 3xdx& =\frac{1}{2}\int \cos (5x-3x)+\cos (5x+3x)dx\\ & =\frac{1}{2}\int \cos (2x)+\cos (8x)dx\\ & =\frac{\sin 2x}{4}+\frac{\sin 8x}{16}+C.\end{array}$$

Example. Evaluate ${\int }_0^{\pi }\sin 3x\sin xdx$.
Using the third identity with $A=3x$ and $B=x$,

$$\begin{array}{rl}{\int }_0^{\pi }\sin 3x\sin xdx& =\frac{1}{2}{\int }_0^{\pi }\cos (3x-x)-\cos (3x+x)dx\\ & =\frac{1}{2}{\int }_0^{\pi }\cos 2x-\cos 4xdx\\ & =\frac{1}{2}{\left[\frac{\sin 2x}{2}-\frac{\sin 4x}{4}\right]}_0^{\pi }\\ & =\frac{1}{2}(0-0)\\ & =0.\end{array}$$

Therefore, the integral is $0$ — no heavy algebra required. In fact, ${\int }_0^{\pi }\sin mx\sin nxdx=0$ whenever $m$ and $n$ are distinct positive integers (and the same happens for $\cos mx\cos nx$); basically the two waves are "orthogonal" over a full period, which is the starting point of Fourier series in second year. Checking whether the multiple-angle terms vanish at the limits before grinding through the algebra can save you a lot of time in an exam.

Integrating powers of $\tan$ and $\sec$

This section is too easy so imma just paste an image

2.2 Reduction Formulae

For non-obvious trigonometric reductions, you're more likely to just use integration by parts.

A reduction formula expresses an integral $I_n$ that depends on a parameter $n$ in terms of a 'smaller' integral of the same type (usually $I_{n-1}$ or $I_{n-2}$). Once you have one, big powers get knocked down step by step until you land on a base case like $I_0$ or $I_1$, which has to be computed directly. Occasionally an identity alone does the job, as in the first example below; every other derivation in this section needs integration by parts.

Example. Suppose that $I_n={\int }_0^{\pi /4}{\tan }^{n}xdx$ whenever $n\ge 0$. Show that

$$I_n=\frac{1}{n-1}-I_{n-2}\forall n\ge 2,$$

and hence evaluate ${\int }_0^{\pi /4}{\tan }^{6}xdx$.
By the identity ${\tan }^{2}x={\sec }^{2}x-1$,

$$\begin{array}{rl}{I}_n& ={\int }_0^{\pi /4}{\tan }^{n-2}x{\tan }^{2}xdx\\ & ={\int }_0^{\pi /4}{\tan }^{n-2}x({\sec }^{2}x-1)dx\\ & ={\int }_0^{\pi /4}{\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\pi /4}{\tan }^{n-2}xdx\\ & ={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\pi /4}-I_{n-2}\\ & =\frac{1}{n-1}-I_{n-2},\end{array}$$

where the first integral was done by inspection, since it is of the form $\int u^{n-2}du$ with $u=\tan x$. Now apply the formula repeatedly:

$$\begin{array}{rl}{\int }_0^{\pi /4}{\tan }^{6}xdx& =I_6\\ & =\frac{1}{5}-I_4\\ & =\frac{1}{5}-\left(\frac{1}{3}-I_2\right)\\ & =\frac{1}{5}-\frac{1}{3}+\left(\frac{1}{1}-I_0\right)\\ & =\frac{1}{5}-\frac{1}{3}+1-{\int }_0^{\pi /4}dx\\ & =\frac{1}{5}-\frac{1}{3}+1-\frac{\pi }{4}\\ & =\frac{13}{15}-\frac{\pi }{4}.\end{array}$$

Therefore, ${\int }_0^{\pi /4}{\tan }^{6}xdx=\frac{13}{15}-\frac{\pi }{4}$. It is important to remember that the base case $I_0$ must be evaluated directly, since the reduction formula is only valid when $n\ge 2$.

Powers of sine and cosine

$$I_n=-\frac{{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}{I}_{n-2}\forall n\ge 2.$$

Proof. Apply integration by parts with

$$\begin{array}{rlrl}u& ={\sin }^{n-1}x& & v=-\cos x\\ u^{\prime }& =(n-1){\sin }^{n-2}x\cos x& & v^{\prime }=\sin x.\end{array}$$

Then

$$\begin{array}{rl}{I}_n& =\int {\sin }^{n-1}x\sin xdx\\ & =-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}x\cos x\cos xdx\\ & =-{\sin }^{n-1}x\cos x+(n-1)\int {\sin }^{n-2}x(1-{\sin }^{2}x)dx\\ & =-{\sin }^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_n,\end{array}$$

using ${\cos }^{2}x=1-{\sin }^{2}x$ in the third line. Gathering the $I_n$ terms on the left-hand side gives

$$n{I}_n=-{\sin }^{n-1}x\cos x+(n-1)I_{n-2},$$

and dividing both sides by $n$ gives the result. $■$

Basically, you peel one factor of $\sin x$ off to integrate by parts, and the Pythagorean identity smuggles $I_n$ back onto the right-hand side; "solving for the integral" is what produces the formula. It is the same trick used for $\int e^x\cos xdx$. Swapping the roles of $\sin$ and $\cos$ gives the analogous formula

$$\boxed{\int {\cos }^{n}xdx=\frac{{\cos }^{n-1}x\sin x}{n}+\frac{n-1}{n}\int {\cos }^{n-2}xdx}$$

Over $[0,\frac{\pi }{2}]$ the boundary term vanishes at both ends (since $\sin 0=0$ and $\cos \frac{\pi }{2}=0$), so both formulae collapse to

$$\boxed{{\int }_0^{\pi /2}{\sin }^{n}xdx=\frac{n-1}{n}{\int }_0^{\pi /2}{\sin }^{n-2}xdx}$$

and identically for ${\cos }^{n}x$.

Example. Evaluate ${\int }_0^{\pi /2}{\sin }^{5}xdx$.
Applying the boxed formula twice,

$$\begin{array}{rl}{\int }_0^{\pi /2}{\sin }^{5}xdx& =\frac{4}{5}{\int }_0^{\pi /2}{\sin }^{3}xdx\\ & =\frac{4}{5}\cdot \frac{2}{3}{\int }_0^{\pi /2}\sin xdx\\ & =\frac{8}{15}[-\cos x{]}_0^{\pi /2}\\ & =\frac{8}{15}(0-(-1))\\ & =\frac{8}{15}.\end{array}$$

Therefore, ${\int }_0^{\pi /2}{\sin }^{5}xdx=\frac{8}{15}$; notice how the odd power stops at the base case $I_1$, which is integrated directly.

Example. Find $\int {\sin }^{4}xdx$.
The reduction formula gives

$$\begin{array}{rl}{I}_4& =-\frac{{\sin }^{3}x\cos x}{4}+\frac{3}{4}{I}_2\\ I_2& =-\frac{\sin x\cos x}{2}+\frac{1}{2}{I}_0\\ & =-\frac{\sin x\cos x}{2}+\frac{x}{2}.\end{array}$$

Substituting $I_2$ into $I_4$,

$$\begin{array}{rl}\int {\sin }^{4}xdx& =-\frac{{\sin }^{3}x\cos x}{4}+\frac{3}{4}\left(-\frac{\sin x\cos x}{2}+\frac{x}{2}\right)+C\\ & =-\frac{{\sin }^{3}x\cos x}{4}-\frac{3\sin x\cos x}{8}+\frac{3x}{8}+C.\end{array}$$

This is exactly what the double-angle identities of 2.1 would give (up to trig identities), but with far less algebra; for something like ${\sin }^{8}x$ the reduction formula is basically the only sane option.

Two-parameter reduction formulae

Some reduction formulae carry two parameters instead of one.

Proof. For the first formula, apply integration by parts with

$$\begin{array}{rlrl}u& ={\cos }^{m-1}x& & v=\frac{{\sin }^{n+1}x}{n+1}\\ u^{\prime }& =-(m-1){\cos }^{m-2}x\sin x& & v^{\prime }={\sin }^{n}x\cos x.\end{array}$$

Then

$$\begin{array}{rl}{I}_{m,n}& ={\left[\frac{{\cos }^{m-1}x{\sin }^{n+1}x}{n+1}\right]}_0^{\pi /2}+\frac{m-1}{n+1}{\int }_0^{\pi /2}{\cos }^{m-2}x{\sin }^{n+2}xdx\\ & =\frac{m-1}{n+1}{\int }_0^{\pi /2}{\cos }^{m-2}x{\sin }^{n}x(1-{\cos }^{2}x)dx\\ & =\frac{m-1}{n+1}{I}_{m-2,n}-\frac{m-1}{n+1}{I}_{m,n},\end{array}$$

since the boundary term vanishes at both ends and ${\sin }^{2}x=1-{\cos }^{2}x$. Gathering the $I_{m,n}$ terms to the left (the same trick as the sine formula) gives

$$\frac{m+n}{n+1}{I}_{m,n}=\frac{m-1}{n+1}{I}_{m-2,n},$$

and multiplying both sides by $\frac{n+1}{m+n}$ gives the first formula. The second formula is proved similarly. $■$

Example. Evaluate ${\int }_0^{\pi /2}{\cos }^{4}x{\sin }^{6}xdx$.
Use the first formula to reduce $m$, then the second to reduce $n$:

$$\begin{array}{rl}{\int }_0^{\pi /2}{\cos }^{4}x{\sin }^{6}xdx& =I_{4,6}\\ & =\frac{3}{10}{I}_{2,6}\\ & =\frac{3}{10}\cdot \frac{1}{8}{I}_{0,6}\\ & =\frac{3}{10}\cdot \frac{1}{8}\cdot \frac{5}{6}{I}_{0,4}\\ & =\frac{3}{10}\cdot \frac{1}{8}\cdot \frac{5}{6}\cdot \frac{3}{4}{I}_{0,2}\\ & =\frac{3}{10}\cdot \frac{1}{8}\cdot \frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}{I}_{0,0}\\ & =\frac{3}{256}\cdot \frac{\pi }{2}\\ & =\frac{3\pi }{512}.\end{array}$$

Therefore, the integral equals $\frac{3\pi }{512}$. Do not forget that the recursion bottoms out at $I_{0,0}={\int }_0^{\pi /2}dx=\frac{\pi }{2}$, which must be evaluated directly. As a sanity check, the substitution $x\mapsto \frac{\pi }{2}-x$ swaps $\sin$ and $\cos$, so $I_{m,n}=I_{n,m}$; evaluating ${\int }_0^{\pi /2}{\cos }^{6}x{\sin }^{4}xdx$ gives the same $\frac{3\pi }{512}$.

Reduction formulae beyond trigonometry

Nothing about reduction formulae is specifically trigonometric; any integrand with a power of $n$ in it is fair game, and exam questions like using $e^x$ or $\ln x$ here.

Example. Suppose that $I_n={\int }_0^1{x}^n{e}^{-x}dx$. Show that

$$I_n=n{I}_{n-1}-\frac{1}{e}\forall n\ge 1,$$

and hence evaluate ${\int }_0^1{x}^3{e}^{-x}dx$.
Integration by parts with $u=x^n$, $v=-e^{-x}$, $u^{\prime }=n{x}^{n-1}$ and $v^{\prime }=e^{-x}$ gives

$$\begin{array}{rl}{I}_n& =[-x^n{e}^{-x}{]}_0^1+n{\int }_0^1{x}^{n-1}{e}^{-x}dx\\ & =-\frac{1}{e}+n{I}_{n-1},\end{array}$$

as required. The base case is

$$\begin{array}{rl}{I}_0& ={\int }_0^1{e}^{-x}dx\\ & =[-e^{-x}{]}_0^1\\ & =1-\frac{1}{e}.\end{array}$$

Now iterate upwards:

$$\begin{array}{rl}{I}_1& =I_0-\frac{1}{e}=1-\frac{2}{e}\\ I_2& =2I_1-\frac{1}{e}=2-\frac{5}{e}\\ I_3& =3I_2-\frac{1}{e}=6-\frac{16}{e}.\end{array}$$

Therefore, ${\int }_0^1{x}^3{e}^{-x}dx=6-\frac{16}{e}$.

A reduction formula for powers of $\sec$

This one connects back to the $\tan$ and $\sec$ section of 2.1; powers of $\sec$ appear constantly after a $x=a\tan \theta$ substitution (see 2.3), so it is worth knowing how to derive this.

Example. Show that if $I_n=\int {\sec }^{n}xdx$, then

$$I_n=\frac{{\sec }^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}{I}_{n-2}\forall n\ge 2,$$

and hence find $\int {\sec }^{4}xdx$.
Write ${\sec }^{n}x={\sec }^{n-2}x{\sec }^{2}x$ and integrate by parts with

$$\begin{array}{rlrl}u& ={\sec }^{n-2}x& & v=\tan x\\ u^{\prime }& =(n-2){\sec }^{n-2}x\tan x& & v^{\prime }={\sec }^{2}x,\end{array}$$

where $u^{\prime }$ comes from $\frac{d}{dx}\sec x=\sec x\tan x$; differentiating a power of $\sec$ conveniently keeps everything in powers of $\sec$ and $\tan$. Then

$$\begin{array}{rl}{I}_n& ={\sec }^{n-2}x\tan x-(n-2)\int {\sec }^{n-2}x{\tan }^{2}xdx\\ & ={\sec }^{n-2}x\tan x-(n-2)\int {\sec }^{n-2}x({\sec }^{2}x-1)dx\\ & ={\sec }^{n-2}x\tan x-(n-2)I_n+(n-2)I_{n-2}.\end{array}$$

Gathering the $I_n$ terms gives

$$(n-1)I_n={\sec }^{n-2}x\tan x+(n-2)I_{n-2},$$

and dividing by $n-1$ gives the formula. Applying it with $n=4$,

$$\begin{array}{rl}\int {\sec }^{4}xdx& =\frac{{\sec }^{2}x\tan x}{3}+\frac{2}{3}\int {\sec }^{2}xdx\\ & =\frac{{\sec }^{2}x\tan x}{3}+\frac{2\tan x}{3}+C.\end{array}$$

Therefore, $\int {\sec }^{4}xdx=\frac{1}{3}{\sec }^{2}x\tan x+\frac{2}{3}\tan x+C$.

As a short remark: reduction formulae are the key ingredient in a famous proof that $\pi$ is irrational. Assuming $\pi =\frac{p}{q}$ for positive integers $p$ and $q$, one defines $I_n=\frac{q^{2n}}{n!}{\int }_{-\pi /2}^{\pi /2}{\left(\frac{{\pi }^2}{4}-x^2\right)}^n\cos xdx$, proves the reduction formula $I_n=(4n-2)q^2{I}_{n-1}-q^4{\pi }^2{I}_{n-2}$ by two applications of integration by parts, and deduces by induction that every $I_n$ is an integer. But since $\frac{a^n}{n!}\to 0$, we also get $0<I_n<1$ for large $n$ — a contradiction. This is [X] (extension) material, so it is not examinable.

2.3 Trigonometric and Hyperbolic Substitutions

There is no systematic way of finding the right substitution for a general integral; however, integrals containing the square root of a quadratic (i.e. surds of the form $\sqrt{\pm x^2\pm a^2}$) almost always yield to a trigonometric or hyperbolic substitution. The table below tells you which one to try.

Basically, each substitution is chosen so that a Pythagorean-type identity turns the expression under the root into a perfect square:

$$\begin{array}{rlrlr}{a}^2-a^2{\sin }^2\theta =a^2{\cos }^2\theta ,& & a^2+a^2{\tan }^2\theta =a^2{\sec }^2\theta ,& & a^2{\cosh }^2\theta -a^2=a^2{\sinh }^2\theta .\end{array}$$

Whether the trigonometric or hyperbolic option is faster depends on the particular integral, but in general the trigonometric ones are favoured because once the integration is done in $\theta$, it is easier to restate the answer in terms of $x$ (you just draw a right triangle).

Example. Evaluate $\int \sqrt{1-x^2}dx$.
The substitution $x=\sin \theta$, $dx=\cos \theta d\theta$ yields

$$\begin{array}{rl}\int \sqrt{1-x^2}dx& =\int \sqrt{1-{\sin }^2\theta }\cos \theta d\theta \\ & =\int \sqrt{{\cos }^2\theta }\cos \theta d\theta \\ & =\int {\cos }^2\theta d\theta \\ & =\frac{1}{2}\int 1+\cos 2\theta d\theta \\ & =\frac{1}{2}\left(\theta +\frac{\sin 2\theta }{2}\right)+C\\ & =\frac{1}{2}(\theta +\sin \theta \cos \theta )+C,\end{array}$$

using the double angle formulae for $\cos$ and then $\sin$. To state the answer in terms of $x$, draw a right triangle containing the angle $\theta$: since $\sin \theta =\frac{x}{1}$, put $x$ on the side opposite $\theta$ and $1$ on the hypotenuse, so the adjacent side is $\sqrt{1-x^2}$ by Pythagoras. Reading off the triangle, $\theta ={\sin }^{-1}x$ and $\cos \theta =\sqrt{1-x^2}$, hence

$$\int \sqrt{1-x^2}dx=\frac{1}{2}\left({\sin }^{-1}x+x\sqrt{1-x^2}\right)+C.$$

This triangle trick is how you convert back at the end of almost every trigonometric substitution, so make sure you are comfortable with it.

Example. Evaluate $\int \frac{dx}{(4+x^2{)}^{3/2}}$.
Since $(4+x^2{)}^{3/2}=(\sqrt{4+x^2}{)}^3$, this is still a surd of a quadratic in disguise, so we use $x=2\tan \theta$, $dx=2{\sec }^2\theta d\theta$ together with ${\tan }^2\theta +1={\sec }^2\theta$:

$$\begin{array}{rl}\int \frac{dx}{(4+x^2{)}^{3/2}}& =\int \frac{2{\sec }^2\theta d\theta }{{\left(\sqrt{4{\tan }^2\theta +4}\right)}^3}\\ & =\int \frac{2{\sec }^2\theta d\theta }{(2\sec \theta {)}^3}\\ & =\frac{1}{4}\int \frac{d\theta }{\sec \theta }\\ & =\frac{1}{4}\int \cos \theta d\theta \\ & =\frac{\sin \theta }{4}+C.\end{array}$$

Now the triangle: $\tan \theta =\frac{x}{2}$, so put $x$ opposite, $2$ adjacent, and the hypotenuse is $\sqrt{x^2+4}$. Thus $\sin \theta =\frac{x}{\sqrt{x^2+4}}$ and

$$\int \frac{dx}{(4+x^2{)}^{3/2}}=\frac{x}{4\sqrt{x^2+4}}+C.$$

Example. Use the substitution $x=3\cosh \theta$ to evaluate $\int \frac{x^3}{\sqrt{x^2-9}}dx$.
With $x=3\cosh \theta$, $dx=3\sinh \theta d\theta$, the identity ${\cosh }^2\theta -{\sinh }^2\theta =1$ gives $\sqrt{x^2-9}=\sqrt{9{\cosh }^2\theta -9}=3\sinh \theta$. Hence

$$\begin{array}{rl}\int \frac{x^3}{\sqrt{x^2-9}}dx& =\int \frac{27{\cosh }^3\theta \cdot 3\sinh \theta d\theta }{3\sinh \theta }\\ & =27\int {\cosh }^3\theta d\theta \\ & =27\int \cosh \theta (1+{\sinh }^2\theta )d\theta \\ & =27\int 1+u^{2}du\\ & =27\left(u+\frac{u^3}{3}\right)+C,\end{array}$$

where the odd power of $\cosh$ was handled with the substitution $u=\sinh \theta$, $du=\cosh \theta d\theta$ — exactly like odd powers of $\cos$ back in 2.1. Since $u=\sinh \theta =\frac{1}{3}\sqrt{x^2-9}$,

$$\begin{array}{rl}\int \frac{x^3}{\sqrt{x^2-9}}dx& =27\left(\frac{\sqrt{x^2-9}}{3}+\frac{(x^2-9{)}^{3/2}}{81}\right)+C\\ & =9\sqrt{x^2-9}+\frac{1}{3}(x^2-9{)}^{3/2}+C.\end{array}$$

Notice how with hyperbolic substitutions you convert back to $x$ through the identity itself ($\sinh \theta =\frac{1}{3}\sqrt{x^2-9}$), no triangle needed. The same integral can also be done with $x=3\sec \theta$; it is a good exercise to check you get the same answer.

Example. Evaluate $\int \frac{dx}{\sqrt{x^2-6x+13}}$.
The quadratic does not match the table until we complete the square:

$$x^2-6x+13=(x-3{)}^2+4.$$

This is now the $\sqrt{a^2+x^2}$ pattern (with $x-3$ playing the role of $x$), so substitute $x-3=2\sinh \theta$, $dx=2\cosh \theta d\theta$:

$$\begin{array}{rl}\int \frac{dx}{\sqrt{(x-3{)}^2+4}}& =\int \frac{2\cosh \theta d\theta }{\sqrt{4{\sinh }^2\theta +4}}\\ & =\int \frac{2\cosh \theta d\theta }{2\cosh \theta }\\ & =\int d\theta \\ & =\theta +C\\ & ={\sinh }^{-1}\left(\frac{x-3}{2}\right)+C.\end{array}$$

Using ${\sinh }^{-1}t=\ln \left(t+\sqrt{t^2+1}\right)$, the answer can also be written as $\ln \left(x-3+\sqrt{x^2-6x+13}\right)+K$. This is a case where the hyperbolic option beats the trigonometric one: $x-3=2\tan \theta$ leads to $\int \sec \theta d\theta$, which is doable but uglier. If the quadratic under the root has a linear term, always complete the square first; none of the standard substitutions apply until you do.

Example. Evaluate ${\int }_0^1\frac{x^2}{\sqrt{4-x^2}}dx$.
Substitute $x=2\sin \theta$, $dx=2\cos \theta d\theta$, and convert the limits: when $x=0$, $\theta =0$; when $x=1$, $\sin \theta =\frac{1}{2}$ so $\theta =\frac{\pi }{6}$. Then

$$\begin{array}{rl}{\int }_0^1\frac{x^2}{\sqrt{4-x^2}}dx& ={\int }_0^{\pi /6}\frac{4{\sin }^2\theta \cdot 2\cos \theta d\theta }{2\cos \theta }\\ & =4{\int }_0^{\pi /6}{\sin }^2\theta d\theta \\ & =2{\int }_0^{\pi /6}1-\cos 2\theta d\theta \\ & =2{\left[\theta -\frac{\sin 2\theta }{2}\right]}_0^{\pi /6}\\ & =2\left(\frac{\pi }{6}-\frac{1}{2}\sin \frac{\pi }{3}\right)\\ & =\frac{\pi }{3}-\frac{\sqrt{3}}{2}.\end{array}$$

Therefore, the integral equals $\frac{\pi }{3}-\frac{\sqrt{3}}{2}$. When substituting in a definite integral, convert the limits of integration as well; then there is no need to convert the antiderivative back to $x$ at all.

2.4 Integrating Rational Functions

The punchline of this section: every rational function has an antiderivative among the elementary functions, and there is a completely systematic procedure for finding it. First, some terminology.

$$f(x)=\frac{p(x)}{q(x)},$$

where $p$ and $q$ are polynomials. We say that $f$ is proper if $\mathrm{deg}p<\mathrm{deg}q$, and improper if $\mathrm{deg}p\ge \mathrm{deg}q$. A quadratic polynomial is irreducible if it has no real linear factors; equivalently, $a{x}^2+bx+c$ is irreducible if its discriminant $b^2-4ac$ is negative.

Basically, proper means bottom-heavy (like $\frac{2}{7}$) and improper means top-heavy (like $\frac{9}{7}$); just as $\frac{9}{7}=1+\frac{2}{7}$, an improper rational function splits into a polynomial plus a proper part via polynomial division. Irreducible quadratics are the ones that refuse to factorise over the reals, like $x^2+4$ or $x^2+x+1$; they are the reason ${\tan }^{-1}$ shows up in the answers.

Before the general strategy, a warm-up that revises the two key tactics for irreducible quadratic denominators.

Example. Evaluate $\int \frac{x}{x^2+2x+10}dx$.
The first tactic is to force the derivative of the denominator ($2x+2$) onto the numerator:

$$\begin{array}{rl}\int \frac{x}{x^2+2x+10}dx& =\frac{1}{2}\int \frac{2x}{x^2+2x+10}dx\\ & =\frac{1}{2}\int \frac{(2x+2)-2}{x^2+2x+10}dx\\ & =\frac{1}{2}\int \frac{2x+2}{x^2+2x+10}dx-\int \frac{dx}{x^2+2x+10}.\end{array}$$

The first integral is of the form $\int \frac{g^{\prime }}{g}$ and becomes a $\ln$. For the second, the second tactic: complete the square in the denominator,

$$x^2+2x+10=(x+1{)}^2+3^2.$$

Hence

$$\begin{array}{rl}\int \frac{x}{x^2+2x+10}dx& =\frac{1}{2}\int \frac{2x+2}{x^2+2x+10}dx-\int \frac{dx}{(x+1{)}^2+3^2}\\ & =\frac{1}{2}\ln (x^2+2x+10)-\frac{1}{3}{\tan }^{-1}\left(\frac{x+1}{3}\right)+C.\end{array}$$

Every proper rational function with an irreducible quadratic denominator is integrated exactly like this; part of it becomes a $\ln$ and the rest becomes a ${\tan }^{-1}$.

The overall strategy

1. If the rational function is improper, use polynomial division to write $f$ as a polynomial plus a proper rational function. The polynomial is trivial to integrate, so we only ever need to worry about proper rational functions.
2. Every proper rational function can be written as a unique sum of functions of the form

$$\begin{array}{rlrlr}\frac{A}{(x-a{)}^k}& & \text{ and }& & \frac{Bx+C}{(x^2+bx+c{)}^k},\end{array}$$

where $x^2+bx+c$ is irreducible. This sum is called the partial fractions decomposition of $f$.
3. Integrate each term of the decomposition using (after completing the square or a simple substitution) the standard formulae

$$\begin{array}{rl}& \boxed{\int x^{k}dx=\frac{x^{k+1}}{k+1}+C,k\ne -1}\\ & \boxed{\int \frac{g^{\prime }(x)}{g(x)}dx=\ln |g(x)|+C}\\ & \boxed{\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C}\end{array}$$

Do not forget the factor of $\frac{1}{a}$ in front of ${\tan }^{-1}\frac{x}{a}$; it is probably the single most commonly dropped term in this whole topic.

Partial fraction decompositions

To find the decomposition of a proper rational function $\frac{p}{q}$, factorise the denominator $q$ as far as possible into real linear factors and real irreducible quadratic factors; the factorisation determines the form of the decomposition, and then the constants are found by algebra. There are a few cases.

Case 1: distinct linear factors. Each factor contributes one term with a constant on top:

$$\begin{array}{rlr}\frac{x-3}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2},& & \frac{x^2-x+7}{x(2x+1)(x-3)}=\frac{A}{x}+\frac{B}{2x+1}+\frac{C}{x-3}.\end{array}$$

Example. Find the partial fractions decomposition of $\frac{7x-1}{x^2-2x-3}$.
Factorising, $x^2-2x-3=(x-3)(x+1)$, so the decomposition takes the form

$$\frac{7x-1}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}.$$

Multiplying through by $(x-3)(x+1)$ gives the polynomial identity

$$7x-1=A(x+1)+B(x-3)\forall x\in \mathbb{R}.$$

Since this holds for all $x$, we may substitute the root of each linear factor to isolate one constant at a time:

$$\begin{array}{rl}x=3& ⟹20=4A⟹A=5\\ x=-1& ⟹-8=-4B⟹B=2.\end{array}$$

Hence

$$\frac{7x-1}{x^2-2x-3}=\frac{5}{x-3}+\frac{2}{x+1},$$

which you can (and should) verify by putting the right-hand side back over a common denominator.

Case 2: repeated linear factors. A factor $(x-a{)}^k$ contributes $k$ terms, one for each power up to $k$:

$$\begin{array}{rlr}\frac{x^2+1}{(x+4{)}^3}=\frac{A}{x+4}+\frac{B}{(x+4{)}^2}+\frac{C}{(x+4{)}^3},& & \frac{x^2-2}{(x-1)(x-2{)}^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2{)}^2}.\end{array}$$

Note carefully that every power of the repeated factor appears in the decomposition, not just the highest one.

Example. Find the partial fractions decomposition of $\frac{x^2-3x+8}{x(x-2{)}^2}$.
The decomposition takes the form

$$\frac{x^2-3x+8}{x(x-2{)}^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2{)}^2}.$$

Multiplying through by $x(x-2{)}^2$,

$$x^2-3x+8=A(x-2{)}^2+Bx(x-2)+Cx\forall x\in \mathbb{R}.$$

The roots determine $A$ and $C$:

$$\begin{array}{rl}x=2& ⟹6=2C⟹C=3\\ x=0& ⟹8=4A⟹A=2.\end{array}$$

There is no third root to substitute, so pick any other small integer (or compare coefficients of $x^2$):

$$\begin{array}{rl}x=1& ⟹6=A-B+C⟹B=A+C-6=-1.\end{array}$$

Hence

$$\frac{x^2-3x+8}{x(x-2{)}^2}=\frac{2}{x}-\frac{1}{x-2}+\frac{3}{(x-2{)}^2}.$$

Case 3: irreducible quadratic factors. Each irreducible quadratic contributes a term with a linear numerator:

$$\begin{array}{rlr}\frac{x^2+x}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9},& & \frac{x^3-2x+4}{(x^2+5)(x^2+x+1)}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{x^2+x+1}.\end{array}$$

The numerator over an irreducible quadratic is $Bx+C$, not just a constant; forgetting the $Bx$ is a classic mistake.

Example. Find the partial fractions decomposition of $\frac{4x^2+2x+1}{(x+1)(x^2+x+1)}$.
The quadratic $x^2+x+1$ has discriminant $1-4=-3<0$, so it is irreducible and the decomposition takes the form

$$\frac{4x^2+2x+1}{(x+1)(x^2+x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+x+1}.$$