Answer
$=\frac{x}{16\sqrt {16-x^{2}}}$
Work Step by Step
$\int \frac{1}{(16-x^{2})^\frac{3}{2}}dx$
Use $x=4\sin \theta$
$dx=4\cos\theta d\theta$
Use $\sin^{-1}\theta =\frac{x}{4}$ for the triangle.
$=\int \frac{4\cos \theta}{(16-16\sin^{2}\theta)^\frac{3}{2}}d\theta$
$=\int \frac{4\cos \theta}{(16\cos^{2}\theta)^\frac{3}{2}}d\theta$
$=\int \frac{4\cos \theta}{64\cos^{3}\theta}d\theta$
$=\frac{1}{16}\int \sec^{2}\theta d\theta$
$=\frac{1}{16} \tan\theta$
Using the triangle created from before, $\tan \theta=\frac{x}{\sqrt {16-x^{2}}}$
$=\frac{x}{16\sqrt {16-x^{2}}}$
