Question

If $\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2},$ then $\int \sqrt{1-\sin 2x}$ dx is equal to ______________

Answer 1

$\sqrt{1-\sin 2x}$

$=\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x}$

$=\sqrt{{\left(\sin x-\cos x\right)}^2}$

$=\left|\sin x-\cos x\right|$

When $\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2},$

$\sin x>\cos x$

$\Rightarrow \sin x-\cos x>0$

$\Rightarrow \sqrt{1-\sin 2x}=\left|\sin x-\cos x\right|=\sin x-\cos x$

$\therefore \int \sqrt{1-\sin 2x}dx$

$=\int \left(\sin x-\cos x\right)dx$

$=\int \sin xdx-\int \cos xdx$

$=-\cos x-\sin x+C$


If $\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2},$ then $\int \sqrt{1-\sin 2x}dx$ is equal to $\underline{-\cos x-\sin x+C}$.