Integration Techniques

2.1 Trigonometric Integrals#

Sine and cosine#

These integrals are of the form

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

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 $\int{\cos^3{x} \sin^4{x}dx}$
We use the substitution $u = \sin{x}, du = \cos{x} dx$ to yield

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

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 $\int{\cos^6{x} \sin^5{x}dx}$.
This time we use $u = \cos{x}, du = -\sin{x}dx$.

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

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

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

to change the integrals into the form

$$\int{\cos^k{2x}dx}.$$

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

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

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

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

$$\begin{aligned}\int \cos^2 2x \, dx &= \frac{1}{2} \int 1 + \cos 4x \, dx \\&= \frac{x}{2} + \frac{\sin 4x}{8} + C_1.\end{aligned}$$

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

$$u = \sin 2x, \qquad du = 2 \cos 2x \, dx$$

gives

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

Thus we obtain

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

Integrating multiple angles of sine and cosine#

These integrals consist of integrals in the form:

$$\begin{align*} \int{\cos{mx}\sin{nx}dx}, && \int{\cos{mx}\cos{nx}} && \text{ or } &&\int{\sin{mx}\sin{nx}}, \end{align*}$$

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

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

Example. Evaluate $\int{\cos{5x}\cos{3x}dx.}$
Using the identity,

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

Example. Evaluate $\int_{0}^{\pi}{\sin{3x}\sin{x}\,dx}$.
Using the third identity with $A = 3x$ and $B = x$,

$$\begin{align*} \int_{0}^{\pi}{\sin{3x}\sin{x}\,dx} &= \frac{1}{2}\int_{0}^{\pi}{\cos(3x-x) - \cos(3x+x)\,dx} \\ &= \frac{1}{2}\int_{0}^{\pi}{\cos{2x} - \cos{4x}\,dx} \\ &= \frac{1}{2}\left[\frac{\sin{2x}}{2} - \frac{\sin{4x}}{4}\right]_{0}^{\pi} \\ &= \frac{1}{2}(0 - 0) \\ &= 0. \end{align*}$$

Therefore, the integral is $0$ — no heavy algebra required. In fact, $\int_{0}^{\pi}{\sin{mx}\sin{nx}\,dx} = 0$ 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 $I_n = \int_{0}^{\pi/4}{\tan^n{x}\,dx}$ whenever $n \geq 0$. Show that

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

and hence evaluate $\int_{0}^{\pi/4}{\tan^6{x}\,dx}$.
By the identity $\tan^2{x} = \sec^2{x} - 1$,

$$\begin{align*} I_n &= \int_{0}^{\pi/4}{\tan^{n-2}{x}\tan^2{x}\,dx} \\ &= \int_{0}^{\pi/4}{\tan^{n-2}{x}(\sec^2{x}-1)\,dx} \\ &= \int_{0}^{\pi/4}{\tan^{n-2}{x}\sec^2{x}\,dx} - \int_{0}^{\pi/4}{\tan^{n-2}{x}\,dx} \\ &= \left[\frac{\tan^{n-1}{x}}{n-1}\right]_{0}^{\pi/4} - I_{n-2} \\ &= \frac{1}{n-1} - I_{n-2}, \end{align*}$$

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:

$$\begin{align*} \int_{0}^{\pi/4}{\tan^6{x}\,dx} &= I_6 \\ &= \frac{1}{5} - I_4 \\ &= \frac{1}{5} - \left(\frac{1}{3} - I_2\right) \\ &= \frac{1}{5} - \frac{1}{3} + \left(\frac{1}{1} - I_0\right) \\ &= \frac{1}{5} - \frac{1}{3} + 1 - \int_{0}^{\pi/4}{dx} \\ &= \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4} \\ &= \frac{13}{15} - \frac{\pi}{4}. \end{align*}$$

Therefore, $\int_{0}^{\pi/4}{\tan^6{x}\,dx} = \frac{13}{15} - \frac{\pi}{4}$. It is important to remember that the base case $I_0$ must be evaluated directly, since the reduction formula is only valid when $n \geq 2$.

Powers of sine and cosine#

$$I_n = -\frac{\sin^{n-1}{x}\cos{x}}{n} + \frac{n-1}{n}I_{n-2} \quad \forall n \geq 2.$$

Proof. Apply integration by parts with

$$\begin{align*} u &= \sin^{n-1}{x} && v = -\cos{x} \\ u' &= (n-1)\sin^{n-2}{x}\cos{x} && v' = \sin{x}. \end{align*}$$

Then

$$\begin{align*} I_n &= \int{\sin^{n-1}{x}\sin{x}\,dx} \\ &= -\sin^{n-1}{x}\cos{x} + (n-1)\int{\sin^{n-2}{x}\cos{x}\cos{x}\,dx} \\ &= -\sin^{n-1}{x}\cos{x} + (n-1)\int{\sin^{n-2}{x}(1-\sin^2{x})\,dx} \\ &= -\sin^{n-1}{x}\cos{x} + (n-1)I_{n-2} - (n-1)I_n, \end{align*}$$

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

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

and dividing both sides by $n$ gives the result. $\blacksquare$

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 $\int{e^x\cos{x}\,dx}$. Swapping the roles of $\sin$ and $\cos$ gives the analogous formula

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

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

$$\boxed{\int_{0}^{\pi/2}{\sin^n{x}\,dx} = \frac{n-1}{n}\int_{0}^{\pi/2}{\sin^{n-2}{x}\,dx}}$$

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

Example. Evaluate $\int_{0}^{\pi/2}{\sin^5{x}\,dx}$.
Applying the boxed formula twice,

$$\begin{align*} \int_{0}^{\pi/2}{\sin^5{x}\,dx} &= \frac{4}{5}\int_{0}^{\pi/2}{\sin^3{x}\,dx} \\ &= \frac{4}{5}\cdot\frac{2}{3}\int_{0}^{\pi/2}{\sin{x}\,dx} \\ &= \frac{8}{15}\Big[-\cos{x}\Big]_{0}^{\pi/2} \\ &= \frac{8}{15}(0-(-1)) \\ &= \frac{8}{15}. \end{align*}$$

Therefore, $\int_{0}^{\pi/2}{\sin^5{x}\,dx} = \frac{8}{15}$; notice how the odd power stops at the base case $I_1$, which is integrated directly.

Example. Find $\int{\sin^4{x}\,dx}$.
The reduction formula gives

$$\begin{align*} I_4 &= -\frac{\sin^3{x}\cos{x}}{4} + \frac{3}{4}I_2 \\ I_2 &= -\frac{\sin{x}\cos{x}}{2} + \frac{1}{2}I_0 \\ &= -\frac{\sin{x}\cos{x}}{2} + \frac{x}{2}. \end{align*}$$

Substituting $I_2$ into $I_4$,

$$\begin{align*} \int{\sin^4{x}\,dx} &= -\frac{\sin^3{x}\cos{x}}{4} + \frac{3}{4}\left(-\frac{\sin{x}\cos{x}}{2} + \frac{x}{2}\right) + C \\ &= -\frac{\sin^3{x}\cos{x}}{4} - \frac{3\sin{x}\cos{x}}{8} + \frac{3x}{8} + C. \end{align*}$$

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

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

Then

$$\begin{align*} I_{m,n} &= \left[\frac{\cos^{m-1}{x}\sin^{n+1}{x}}{n+1}\right]_{0}^{\pi/2} + \frac{m-1}{n+1}\int_{0}^{\pi/2}{\cos^{m-2}{x}\sin^{n+2}{x}\,dx} \\ &= \frac{m-1}{n+1}\int_{0}^{\pi/2}{\cos^{m-2}{x}\sin^{n}{x}(1-\cos^2{x})\,dx} \\ &= \frac{m-1}{n+1}I_{m-2,n} - \frac{m-1}{n+1}I_{m,n}, \end{align*}$$

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. $\blacksquare$

Example. Evaluate $\int_{0}^{\pi/2}{\cos^4{x}\sin^6{x}\,dx}$.
Use the first formula to reduce $m$, then the second to reduce $n$:

$$\begin{align*} \int_{0}^{\pi/2}{\cos^4{x}\sin^6{x}\,dx} &= I_{4,6} \\ &= \frac{3}{10}I_{2,6} \\ &= \frac{3}{10}\cdot\frac{1}{8}I_{0,6} \\ &= \frac{3}{10}\cdot\frac{1}{8}\cdot\frac{5}{6}I_{0,4} \\ &= \frac{3}{10}\cdot\frac{1}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}I_{0,2} \\ &= \frac{3}{10}\cdot\frac{1}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_{0,0} \\ &= \frac{3}{256}\cdot\frac{\pi}{2} \\ &= \frac{3\pi}{512}. \end{align*}$$

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 $\int_{0}^{\pi/2}{\cos^6{x}\sin^4{x}\,dx}$ 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

$$I_n = nI_{n-1} - \frac{1}{e} \quad \forall n \geq 1,$$

and hence evaluate $\int_{0}^{1}{x^3 e^{-x}\,dx}$.
Integration by parts with $u = x^n$, $v = -e^{-x}$, $u' = nx^{n-1}$ and $v' = e^{-x}$ gives

$$\begin{align*} I_n &= \Big[-x^n e^{-x}\Big]_{0}^{1} + n\int_{0}^{1}{x^{n-1}e^{-x}\,dx} \\ &= -\frac{1}{e} + nI_{n-1}, \end{align*}$$

as required. The base case is

$$\begin{align*} I_0 &= \int_{0}^{1}{e^{-x}\,dx} \\ &= \Big[-e^{-x}\Big]_{0}^{1} \\ &= 1 - \frac{1}{e}. \end{align*}$$

Now iterate upwards:

$$\begin{align*} I_1 &= I_0 - \frac{1}{e} = 1 - \frac{2}{e} \\ I_2 &= 2I_1 - \frac{1}{e} = 2 - \frac{5}{e} \\ I_3 &= 3I_2 - \frac{1}{e} = 6 - \frac{16}{e}. \end{align*}$$

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 $I_n = \int{\sec^n{x}\,dx}$, then

$$I_n = \frac{\sec^{n-2}{x}\tan{x}}{n-1} + \frac{n-2}{n-1}I_{n-2} \quad \forall n \geq 2,$$

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

$$\begin{align*} u &= \sec^{n-2}{x} && v = \tan{x} \\ u' &= (n-2)\sec^{n-2}{x}\tan{x} && v' = \sec^2{x}, \end{align*}$$

where $u'$ 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

$$\begin{align*} I_n &= \sec^{n-2}{x}\tan{x} - (n-2)\int{\sec^{n-2}{x}\tan^2{x}\,dx} \\ &= \sec^{n-2}{x}\tan{x} - (n-2)\int{\sec^{n-2}{x}(\sec^2{x}-1)\,dx} \\ &= \sec^{n-2}{x}\tan{x} - (n-2)I_n + (n-2)I_{n-2}. \end{align*}$$

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 $n = 4$,

$$\begin{align*} \int{\sec^4{x}\,dx} &= \frac{\sec^2{x}\tan{x}}{3} + \frac{2}{3}\int{\sec^2{x}\,dx} \\ &= \frac{\sec^2{x}\tan{x}}{3} + \frac{2\tan{x}}{3} + C. \end{align*}$$

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

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 $I_n = \frac{q^{2n}}{n!}\int_{-\pi/2}^{\pi/2}{\left(\frac{\pi^2}{4} - x^2\right)^n \cos{x}\,dx}$, 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:

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

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

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

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 $\tan^2{\theta} + 1 = \sec^2{\theta}$:

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

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

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

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 $u = \sinh{\theta} = \frac{1}{3}\sqrt{x^2-9}$,

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

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 $x - 3 = 2\sinh{\theta}$, $dx = 2\cosh{\theta}\,d\theta$:

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

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

$$\begin{align*} \int_{0}^{1}{\frac{x^2}{\sqrt{4-x^2}}dx} &= \int_{0}^{\pi/6}{\frac{4\sin^2{\theta}\cdot 2\cos{\theta}\,d\theta}{2\cos{\theta}}} \\ &= 4\int_{0}^{\pi/6}{\sin^2{\theta}\,d\theta} \\ &= 2\int_{0}^{\pi/6}{1-\cos{2\theta}\,d\theta} \\ &= 2\left[\theta - \frac{\sin{2\theta}}{2}\right]_{0}^{\pi/6} \\ &= 2\left(\frac{\pi}{6} - \frac{1}{2}\sin{\frac{\pi}{3}}\right) \\ &= \frac{\pi}{3} - \frac{\sqrt{3}}{2}. \end{align*}$$

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.