# Evaluate the integral \int \sqrt{9 - x^2} \, dx using appropriate trigonometric substitution.

## Question:

Evaluate the integral {eq}\int \sqrt{9 - x^2} \, dx {/eq} using appropriate trigonometric substitution.

## Integrations

First we assume u is equal to some value in order to simplify the solution

Then we integrate the given function and substitute back the u value

Formulas used

{eq}\displaystyle \begin{align} \int a\cdot f\left(x\right)dx&=a\cdot \int f\left(x\right)dx\\ \int f\left(x\right)\pm g\left(x\right)dx&=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\ \int \cos \left(u\right)du&=\sin \left(u\right) \end{align} {/eq}

Given

{eq}\displaystyle \int \sqrt{9 - x^2} \, dx {/eq}

Assume

{eq}\displaystyle \begin{align} x&=3\sin(u)\\ dx&=3\cos(u)\\ \end{align} {/eq}

Therefore

{eq}\displaystyle \begin{align} \int \sqrt{9 - x^2} \, dx&=\sqrt{9-9\sin^(x)}3\cos(u)du\\ &=9\cdot \int \cos ^2\left(u\right)du\\ \end{align} {/eq}

We know that

{eq}\displaystyle \begin{align} \cos ^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2} \end{align} {/eq}

Therefore

{eq}\displaystyle \begin{align} \int \sqrt{9 - x^2} \, dx&=9\cdot \frac{1}{2}\cdot \int \:1+\cos \left(2u\right)du\\ &&\int \cos \left(u\right)du=\sin \left(u\right)\\ &=9\cdot \frac{1}{2}\left(\int \:1du+\int \cos \left(2u\right)du\right)+C\\ &=9\cdot \frac{1}{2}\left(u+\frac{1}{2}\sin \left(2u\right)\right)+C\\ \end{align} {/eq}

(Where C is the integrating constant)

Substitute back u value we get

Therefore the final resul is

{eq}\displaystyle \color{blue}{\int \sqrt{9 - x^2} \, dx=\frac{9}{2}\left(\arcsin \left(\frac{1}{3}x\right)+\frac{1}{2}\sin \left(2\arcsin \left(\frac{1}{3}x\right)\right)\right)+C} {/eq}