# Integration by trigonometric substitution

## Redirected from Trigonometric substitution

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Integration by trigonometric substitution is a technique of integration that involves substituting some function of x for a trigonometric function.

As a general rule, when taking an antiderivative of a function in the form $\sqrt{a^2 - x^2}$, the substitution $x=a \sin \theta$ is usually the best option. For $\sqrt{x^2 + a^2}$ and $\sqrt{x^2 - a^2}$, the substitutions $x=a \tan \theta$ and $x=a \sec \theta$ (respectively) are usually the best options.

## Examples

This technique can be used when functions would be otherwise difficult to integrate. One of the most well-known examples is $\int \frac{1}{\sqrt{1-x^2}} dx$

Here, we can use the substitution $x=\sin{\theta}, \ dx= \cos{\theta} \ d\theta$ to get

$\int \frac{\cos \theta}{\sqrt{1-\sin^2 \theta}} \ d \theta = \int \frac{\cos \theta}{\sqrt{\cos^2 \theta}} \ d \theta = \int d \theta = \theta$

$x= \sin{\theta} \Rightarrow \arcsin x = \theta$

Therefore:

$\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x$

A second example:

$\int \frac{1}{x^2 \sqrt{4+x^2}} dx = \int \frac{1}{x^2 \sqrt{2^2+x^2}} dx$

Here, we can use the substitution $x=2 \tan{\theta}, \ dx=2 \sec^2{\theta}d\theta$ to get

$\int \frac{ 2 \sec^2{\theta} }{(2 \tan{\theta})^2 \sqrt{4(1 + \tan^2 \theta)}} d\theta = \int \frac{ 2 \sec^2{\theta} }{ 8 \tan^2 {\theta} \sqrt{1 + \tan^2 \theta}} d\theta = \frac{1}{4} \int \frac{\sec^2 \theta }{\tan^2 {\theta} \sqrt{1 + \tan^2 \theta}} d\theta$

By using the trigonometric identity $\tan^2 \theta + 1 = \sec^2 \theta$, we get

$\frac{1}{4} \int \frac{\sec^2 \theta }{\tan^2 {\theta} \sqrt{\sec^2 \theta}} d\theta = \frac{1}{4} \int \frac{\sec^2 \theta }{\tan^2 {\theta} \sec \theta} d\theta = \frac{1}{4} \int \frac{\sec \theta}{\tan^2 {\theta}} d\theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 {\theta}} d\theta$

Which evaluates to $\frac{-1}{\sin \theta}$ by using u-substitution. Since

$x=2 \tan{\theta} \Rightarrow \arctan \tfrac{x}{2} = \theta$

we can say that

$\frac{-1}{4 \sin \theta} = \frac{-1}{4 \sin (\arctan \tfrac{x}{2})} = \frac{-1}{4 \tfrac{x}{\sqrt{x^2 + 4}}} = -\frac{\sqrt{x^2 + 4}}{4x}$

Therefore:

$\int \frac{1}{x^2 \sqrt{4+x^2}} dx = -\frac{\sqrt{x^2 + 4}}{4x} + C$