# How do you find the integral of (x^2)/(16-x^2)^(1/2)?

Apr 20, 2018

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

#### Explanation:

Rewrite with a root:

$\int {x}^{2} / \sqrt{16 - {x}^{2}} \mathrm{dx}$

For integrals involving the root $\sqrt{{a}^{2} - {x}^{2}} ,$ we use the substitution $x = a \sin \theta .$

Here, ${a}^{2} = 16 , a = 4 , x = 4 \sin \theta , \mathrm{dx} = 4 \cos \theta d \theta$ and we get

$\int \frac{16 {\sin}^{2} \theta 4 \cos \theta d \theta}{\sqrt{16 \left(1 - {\sin}^{2} \theta\right)}}$

Recalling that $1 - {\sin}^{2} \theta = {\cos}^{2} \theta ,$we get

$= 16 \int \frac{{\sin}^{2} \theta \cos \theta d \theta}{\sqrt{{\cos}^{2} \theta}}$

$= 16 \int \frac{{\sin}^{2} \theta \cancel{\cos} \theta}{\cancel{\cos} \theta} d \theta$

$= 16 \int {\sin}^{2} \theta d \theta$

Recalling that ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right) ,$ we get

$\frac{16}{2} \int \left(1 - \cos 2 \theta d \theta\right) = 8 \left(\theta - \frac{1}{2} \sin 2 \theta\right) + C$

We need to get things in terms of $x$.

Recalling that $x = 4 \sin \theta , \sin \theta = \frac{x}{4} , \theta = \arcsin \left(\frac{x}{4}\right)$

To determine $\frac{1}{2} \sin 2 \theta ,$ recall the identity $\frac{1}{2} \sin 2 \theta = \sin \theta \cos \theta .$ To use this, we need to determine the cosine using the below Pythagorean identity:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$
${\cos}^{2} \theta = 1 - {x}^{2} / 16$

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

Thus, $\frac{1}{2} \sin 2 \theta = \left(\frac{x}{4}\right) \frac{\sqrt{16 - {x}^{2}}}{4} = \frac{x \sqrt{16 - {x}^{2}}}{16}$

And

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