Question

The value of the integral $\sqrt{2}{\int }_0^{100\pi }\sqrt{1-\cos 2x}dx$ is

Answer 1

We have $\sqrt{1-\cos 2x}=\sqrt{2}\left|\sin x\right|$
Therefore, since $\left|\sin x\right|$has the period $\pi$, we have
${\int }_0^{100\pi }\sqrt{1-\cos 2x}dx=\sqrt{2}{\int }_0^{100\pi }\left|\sin x\right|dx$
$=100\sqrt{2}{\int }_0^{\pi }\sin xdx=200\sqrt{2}$