Answer
$$\int^{\pi}_0\sqrt{1-\cos2x}dx=2\sqrt2$$
Work Step by Step
$$A=\int^{\pi}_0\sqrt{1-\cos2x}dx$$
Use the identity $$\frac{1-\cos 2x}{2}=\sin^2x$$ $$1-\cos2x=2\sin^2x$$
That means $$A=\int^{\pi}_0\sqrt{2\sin^2x}dx$$ $$A=\sqrt2\int^{\pi}_0|\sin x|dx$$
As $\sin x\ge0$ on $[0,\pi]$, $|\sin x|=\sin x$
$$A=\sqrt2\int^{\pi}_0\sin xdx$$ $$A=-\sqrt2\cos x\Big]^{\pi}_0$$ $$A=-\sqrt2(\cos\pi-\cos0)$$ $$A=-\sqrt2(-1-1)=2\sqrt2$$
