Using $x=4\sin\theta$, where $0

Question

Using $x=4\sin\theta$, where $0<\theta<\dfrac{\pi}{2}$, find $\displaystyle\int\dfrac{\sqrt{16-x^2}}{x^2}\,\mathrm{d}x$ for $0<x<4$.

Accepted Answer

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

A Supplied Sine Substitution