# How do you integrate int x^2sqrt(16-x^2) by trigonometric substitution?

Mar 23, 2018

The answer is $= 32 \arcsin \left(\frac{x}{4}\right) - 2 \sqrt{16 - {x}^{2}} \left(x - \frac{1}{8} {x}^{3}\right) + C$

#### Explanation:

Perform the substitution

$x = 4 \sin \theta$, $\implies$, $\mathrm{dx} = 4 \cos \theta d \theta$

$\sqrt{16 - {x}^{2}} = \sqrt{16 - 16 {\sin}^{2} \theta} = 4 \cos \theta$

Therefore,

The integral is

$I = \int {x}^{2} \sqrt{16 - {x}^{2}} \mathrm{dx} = \int 16 {\sin}^{2} \theta \cdot 4 \cos \theta \cdot 4 \cos \theta d \theta$

$= 256 \int {\sin}^{2} \theta {\cos}^{2} \theta d \theta$

$2 \sin \theta \cos \theta = \sin \left(2 \theta\right)$

$I = 64 \int {\sin}^{2} \left(2 \theta\right) d \theta$

$\cos \left(4 \theta\right) = 1 - 2 {\sin}^{2} \left(2 \theta\right)$

${\sin}^{2} \left(\theta\right) = \frac{1 - \cos \left(4 \theta\right)}{2}$

Therefore,

$I = 32 \int \left(1 - \cos \left(4 \theta\right)\right) d \theta$

$= 32 \left(\theta - \sin \frac{4 \theta}{4}\right)$

$= 32 \arcsin \left(\frac{x}{4}\right) - 8 \sin \left(4 \theta\right)$

$\sin \left(4 \theta\right) = \cos \theta \left(4 \sin \theta - 8 {\sin}^{3} \theta\right)$

$= \frac{\sqrt{16 - {x}^{2}}}{4} \left(x - \frac{1}{8} {x}^{3}\right)$

And finally,

$\int {x}^{2} \sqrt{16 - {x}^{2}} \mathrm{dx} = 32 \arcsin \left(\frac{x}{4}\right) - 2 \sqrt{16 - {x}^{2}} \left(x - \frac{1}{8} {x}^{3}\right) + C$