Question

Evaluate: ${\int }_0^{\pi /2}\sqrt{1-\cos 2x}dx$.

Answer 1

${\int }_0^{\frac{\pi }{2}}\sqrt{1-\cos 2xdx}$

$={\int }_0^{\frac{\pi }{2}}\sqrt{1-1+2{\sin }^{2}x}dx$

$=\sqrt{2}{\int }_0^{\frac{\pi }{2}}\sin xdx$

$=-\sqrt{2}{\left[\cos x\right]}_0^{\pi /2}$

$=-\sqrt{2}\left(0-1\right)$

$I=\sqrt{2}$

Hence, this is the answer.