# How do you findint (x^2) / (sqrt(81 - x^2)) dx  using trigonometric substitution?

Dec 6, 2015

$= \frac{81}{2} {\sin}^{- 1} \left(\frac{x}{9}\right) - \frac{x \sqrt{81 - {x}^{2}}}{2} + C$

#### Explanation:

Let $x = 9 \sin \theta$
Then $\mathrm{dx} = 9 \cos \theta d \theta$ and $\sin \theta = \frac{x}{9}$

Then the original integral becomes

$9 \int \frac{81 {\sin}^{2} \theta}{\sqrt{81 - 81 {\sin}^{2} \theta}} \cos \theta d \theta$

Now using the trig identities ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ and
$\sin 2 \theta = 2 \sin \theta \cos \theta$, together with factorizations and laws of surds, we may compute the integral as

$9 \times 81 \int \frac{{\sin}^{2} \theta}{9 \sqrt{1 - {\sin}^{2} \theta}} \cos \theta d \theta$

$= 81 \int \frac{{\sin}^{2} \theta}{\cos} \theta \cos \theta d \theta$

$= 81 \left[\frac{\theta}{2} - \frac{\sin 2 \theta}{4}\right] + C$

$= \frac{81}{2} {\sin}^{- 1} \left(\frac{x}{9}\right) - \frac{81}{2} \cdot \frac{x}{9} \cdot \frac{\sqrt{81 - {x}^{2}}}{9} + C$