# How do you integrate int (6-x^2)/sqrt(9-x^2) dx using trigonometric substitution?

Nov 8, 2016

$\frac{1}{2} x \sqrt{9 - {x}^{2}} + \frac{3}{2} \arcsin \left(\frac{x}{3}\right) + C$

#### Explanation:

$I = \int \frac{6 - {x}^{2}}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

Rewrite the integral for simplification:

$I = \int \frac{- 3 + 9 - {x}^{2}}{\sqrt{9 - {x}^{2}}} \mathrm{dx} = - 3 \int \frac{\mathrm{dx}}{\sqrt{9 - {x}^{2}}} + \int \frac{9 - {x}^{2}}{\sqrt{9 - {x}^{2}}} \mathrm{dx}$

$\textcolor{w h i t e}{I} = - 3 \int \frac{\mathrm{dx}}{\sqrt{9 - {x}^{2}}} + \int \sqrt{9 - {x}^{2}} \mathrm{dx}$

First focusing on $J = \int \frac{\mathrm{dx}}{\sqrt{9 - {x}^{2}}}$. For this, let $x = 3 \sin \theta$. This implies that $\mathrm{dx} = 3 \cos \theta d \theta$.

$J = \int \frac{\mathrm{dx}}{\sqrt{9 - {x}^{2}}} = \int \frac{3 \cos \theta d \theta}{\sqrt{9 - 9 {\sin}^{2} \theta}} = \int \cos \frac{\theta}{\sqrt{1 - {\sin}^{2} \theta}} d \theta = \int d \theta$

$\textcolor{w h i t e}{J} = \theta = \arcsin \left(\frac{x}{3}\right)$

Recall that the $\theta = \arcsin \left(\frac{x}{3}\right)$ relationship comes from our earlier substitution $x = 3 \sin \theta$.

Now we can work on the next integral $K = \int \sqrt{9 - {x}^{2}} \mathrm{dx}$. We will use the same substitution as before, $x = 3 \sin \phi$, implying that $\mathrm{dx} = 3 \cos \phi \mathrm{dp} h i$.

$K = \int \sqrt{9 - {x}^{2}} \mathrm{dx} = \int \sqrt{9 - 9 {\sin}^{2} \phi} \left(3 \cos \phi \mathrm{dp} h i\right)$

$\textcolor{w h i t e}{K} = 9 \int \sqrt{1 - {\sin}^{2} \phi} \left(\cos \phi\right) \mathrm{dp} h i = 9 \int {\cos}^{2} \phi \mathrm{dp} h i$

Rewrite by solving for ${\cos}^{2} x$ in the identity $\cos 2 x = 2 {\cos}^{2} x - 1$.

$K = 9 \int \frac{\cos 2 \phi + 1}{2} \mathrm{dp} h i = \frac{9}{2} \int \cos 2 \phi \mathrm{dp} h i + \frac{9}{2} \int \mathrm{dp} h i$

For the first integral, we can use the substitution $u = 2 \phi$ so $\mathrm{du} = 2 \mathrm{dp} h i$ and $\mathrm{dp} h i = \frac{1}{2} \mathrm{du}$. The second integral is a fundamental one:

$K = \frac{9}{4} \int \cos u \mathrm{du} + \frac{9}{2} \phi = \frac{9}{4} \sin u + \frac{9}{2} \phi = \frac{9}{4} \sin 2 \phi + \frac{9}{2} \phi$

Using $\sin 2 x = 2 \sin x \cos x$:

$K = \frac{9}{2} \sin \phi \cos \phi + \frac{9}{2} \phi$

Now to undo these substitutions, recall that $x = 3 \sin \phi$. Thus, $\phi = \arcsin \left(\frac{x}{3}\right)$ and $\sin \phi = \frac{x}{3}$. Thus $\cos \phi = \sqrt{1 - {\sin}^{2} \phi} = \sqrt{1 - {x}^{2} / 9} = \frac{1}{3} \sqrt{9 - {x}^{2}}$. So:

$K = \frac{9}{2} \left(\frac{x}{3}\right) \left(\frac{1}{3} \sqrt{9 - {x}^{2}}\right) + \frac{9}{2} \arcsin \left(\frac{x}{3}\right)$

$\textcolor{w h i t e}{K} = \frac{1}{2} x \sqrt{9 - {x}^{2}} + \frac{9}{2} \arcsin \left(\frac{x}{3}\right)$

Returning to the original integral:

$I = - 3 J + K$

$\textcolor{w h i t e}{I} = - 3 \arcsin \left(\frac{x}{3}\right) + \left[\frac{1}{2} x \sqrt{9 - {x}^{2}} + \frac{9}{2} \arcsin \left(\frac{x}{3}\right)\right]$

$\textcolor{w h i t e}{I} = \frac{1}{2} x \sqrt{9 - {x}^{2}} + \frac{3}{2} \arcsin \left(\frac{x}{3}\right) + C$