## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{\pi/2}_0\theta\sqrt{1-\cos2\theta}d\theta=\sqrt2$$
$$A=\int^{\pi/2}_0\theta\sqrt{1-\cos2\theta}d\theta$$ 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 $$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0-\int^{\pi/2}_0-\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\theta\cos\theta\Big]^{\pi/2}_0+\int^{\pi/2}_0\cos\theta d\theta\Big)$$ $$A=\sqrt2\Big(-\frac{\pi}{2}\cos\frac{\pi}{2}-0+(\sin\theta]^{\pi/2}_0)\Big)$$ $$A=\sqrt2\Big(0+\sin\frac{\pi}{2}-\sin0\Big)$$ $$A=\sqrt2(1)=\sqrt2$$