$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^{\pi / 4} \sqrt{1-\sin 2 x} \mathrm{~d} x=\int_0^{\pi / 4} \sqrt{\cos ^2 x+\sin ^2 x-2 \sin x \cdot \cos x} \mathrm{~d} x \ & =\int_0^{\pi / 4} \sqrt{(\cos x-\sin x)^2} \mathrm{~d} x=\int_0^{\pi / 4}|\cos x-\sin x| \mathrm{d} x \ & =\int_0^{\pi / 4}(\cos x-\sin x) \mathrm{d} x=[\sin x+\cos x]_0^{\pi / 4} \ & =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}ight)-(0+1)=\sqrt{2}-1\end{aligned}$

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} d x=$

Solution