By considering $x=4\cos \theta $ where $0\le \theta \le \pi $, find \(\displaystyle \int{\s

Question

By considering $x=4\cos \theta $ where $0\le \theta \le \pi $, find $\displaystyle \int{\sqrt{16-{{x}^{2}}}\,\mathrm{d}x}$.

Accepted Answer

$-8\cos^{-1}\left(\dfrac{x}{4}\right)+\dfrac{1}{2}x\sqrt{16-x^2}+C$

Supplied Substitutions