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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 $x=a\sin(u)$ is usually the best option. For $\sqrt{a^2+x^2}$ and $\sqrt{a^2-x^2}$ , the substitutions $x=a\tan(u)$ and $x=a\sec(u)$ (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

$\int\dfrac{dx}{\sqrt{1-x^2}}$

Here, we can use the substitution $x=\sin(u),\ dx=\cos(u)du$ to get

\begin{align}&\int\dfrac{\cos(u)}{\sqrt{1-\sin^2(u)}}du=\int\dfrac{\cos(u)}{\sqrt{\cos^2(u)}}du=\int du=u\\&x=\sin(u)\Rightarrow\arcsin(x)=u\end{align}

Therefore:

$\int\dfrac{dx}{\sqrt{1-x^2}}=\arcsin(x)$

A second example:

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

Here, we can use the substitution $x=2\tan(u),\ dx=2\sec^2(u)du$ to get

\begin{align}\int\dfrac{2\sec^2(u)}{(2\tan(u))^2\sqrt{4(1+\tan^2(u))}}du&=\int\dfrac{2\sec^2(u)}{8\tan^2(u)\sqrt{1+\tan^2(u)}}du\\&=\frac14\int\dfrac{\sec^2(u)}{\tan^2(u)\sqrt{1+\tan^2(u)}}du\end{align}

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

\begin{align}\frac14\int\dfrac{\sec^2(u)}{\tan^2(u)\sqrt{\sec^2(u)}}du&=\frac14\int\dfrac{\sec^2(u)}{\tan^2(u)\sec(u)}du\\&=\frac14\int\frac{\sec(u)}{\tan^2(u)}du=\frac14\int\frac{\cos(u)}{\sin^2(u)}du\end{align}

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

$x=2\tan(u)\Rightarrow\arctan\left(\tfrac{x}{2}\right)=u$

we can say that

$-\frac{1}{4\sin(u)}=-\frac{1}{4\sin\left(\arctan\left(\tfrac{x}{2}\right)\right)} =-\frac{1}{\dfrac{4x}{\sqrt{4+x^2}}}=-\frac{\sqrt{4+x^2}}{4x}$

Therefore:

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