Answer
$$\int^{\pi/2}_0\theta\sqrt{1-\cos2\theta}d\theta=\sqrt2$$
Work Step by Step
$$A=\int^{\pi/2}_0\theta\sqrt{1-\cos2\theta}d\theta$$
Use the identity: $$\frac{1-\cos2\theta}{2}=\sin^2\theta$$ $$1-\cos2\theta=2\sin^2\theta$$
Therefore, $$A=\int^{\pi/2}_0\theta\sqrt{2\sin^2\theta}d\theta$$ $$A=\sqrt2\int^{\pi/2}_0\theta\sin\theta d\theta$$ (since on $[0,\pi/2]$, $\sin\theta\ge0$)
Now we apply Integration by Parts.
Set $u=\theta$ and $dv=\sin\theta d\theta$
So we would have $du=d\theta$ and $v=-\cos\theta$
According to Integration by Parts rule, we have
$$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0-\int^{\pi/2}_0-\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0+\int^{\pi/2}_0\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\frac{\pi}{2}\cos\frac{\pi}{2}-0+(\sin\theta]^{\pi/2}_0)\Big)$$ $$A=\sqrt2\Big(0+\sin\frac{\pi}{2}-\sin0\Big)$$ $$A=\sqrt2(1)=\sqrt2$$
