Question

Evaluate ${\int }_0^{200\pi }\sqrt{1+\cos x}dx$

Answer 1

${\int }_0^{200\pi }\sqrt{1+cosx}={\int }_0^{200\pi }\sqrt{1+2{\cos }^2\frac{x}{2}-1}=\sqrt{2}{\int }_0^{200\pi }\left|\cos \frac{x}{2}\right|$