# How do I take the integral of this?

Sep 11, 2017

$\frac{x}{\sqrt{16 - {x}^{2}}} - {\sin}^{-} 1 \left(\frac{x}{4}\right) + C$

#### Explanation:

$\int {x}^{2} / {\left(16 - {x}^{2}\right)}^{\frac{3}{2}} \mathrm{dx}$

Let's use the trigonometric substitution $x = 4 \sin \theta$. This implies that $\mathrm{dx} = 4 \cos \theta d \theta$. Then the integral equals:

$= \int \frac{16 {\sin}^{2} \theta}{16 - 16 {\sin}^{2} \theta} ^ \left(\frac{3}{2}\right) \left(4 \cos \theta d \theta\right)$

$= 64 \int \frac{{\sin}^{2} \theta \cos \theta}{{16}^{\frac{3}{2}} {\left(1 - {\sin}^{2} \theta\right)}^{\frac{3}{2}}} d \theta$

Recall that $1 - {\sin}^{2} \theta = {\cos}^{2} \theta$:

$= \frac{64}{64} \int \frac{{\sin}^{2} \theta \cos \theta}{\cos} ^ 3 \theta d \theta$

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

$= \int {\tan}^{2} \theta d \theta$

Use the identity ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$:

$= \int \left({\sec}^{2} \theta - 1\right) d \theta$

Both of these terms can be integrated easily:

$= \tan \theta - \theta + C$

From our substitution $x = 4 \sin \theta$, we can rearrange to find that $\theta = {\sin}^{-} 1 \left(\frac{x}{4}\right)$.

Also, since $\sin \theta = \frac{x}{4}$, we have a right triangle where the side opposite $\theta$ is $x$ and the hypotenuse is $4$. From the Pythagorean Theorem, we find that the side adjacent to $\theta$ is $\sqrt{16 - {x}^{2}}$. Thus, $\tan \theta = \frac{x}{\sqrt{16 - {x}^{2}}}$. The integral then equals:

$= \frac{x}{\sqrt{16 - {x}^{2}}} - {\sin}^{-} 1 \left(\frac{x}{4}\right) + C$