Question

Integrate:

${\int }_0^{\pi /4}\cos 2x\sqrt{1-\sin 2x}dx$

Answer 1

let $\sin 2x=t$

$dx.2\cos 2x=dt$

$I=\frac{1}{2}{\int }_0^{\pi /2}2\cos 2x\sqrt{1-\sin 2xdx}$

$=\frac{1}{2}{\int }_0^{\pi /4}dt\sqrt{1-t}=\frac{1}{2}{\left[\frac{(1-t{)}^{1/2+1}}{\frac{1}{2}+1}\right]}_0^{\pi /4}$

$={\left[\frac{(1-t{)}^{3/2}}{3}\right]}_0^{\pi /4}=\frac{1}{3}[1-\sin 2x{]}_0^{\pi /4}$

$=\frac{1}{3}[1-\sin 90]-[1-0]=\frac{-1}{3}$