Question

If 0 < x < $\frac{\mathrm{\pi }}{4}$, 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 $0<x<\frac{\mathrm{\pi }}{4},$

$\sin x<\cos x$

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

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

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

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

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

$=-\left(-\cos x\right)+\sin x+C$

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


If 0 < x < $\frac{\mathrm{\pi }}{4}$, then $\int \sqrt{1-\sin 2x}dx$ is equal to $\underline{\cos x+\sin x+C}$.