Answer
\begin{aligned}
\int_{0}^{\pi} \sqrt{1-\cos 2 x} d x =2 \sqrt{2}
\end{aligned}
Work Step by Step
Given $$\int_{0}^{\pi} \sqrt{1-\cos 2 x} d x $$
So, we have
\begin{aligned}
I&=\int_{0}^{\pi} \sqrt{1-\cos 2 x} d x\\
&\text{since} \ \ \cos2 x= 1-2\sin^2x \Rightarrow2 \sin ^{2} x=1-\cos x,\text{ we get}\\
I&=\int_{0}^{\pi} \sqrt{2}|\sin x| d x\\
&=\int_{0}^{\pi} \sqrt{2} \sin x d x\\
&=[-\sqrt{2} \cos x]_{0}^{\pi}\\
&=[-\sqrt{2} \cos \pi]-[-\sqrt{2} \cos 0]\\
&=\sqrt{2}+\sqrt{2}\\
&=2 \sqrt{2}
\end{aligned}
