University Calculus: Early Transcendentals (3rd Edition)

$$\int^{2\pi}_0\sqrt{\frac{1-\cos x}{2}}dx=4$$
$$A=\int^{2\pi}_0\sqrt{\frac{1-\cos x}{2}}dx$$ Use the identity $$\frac{1-\cos 2x}{2}=\sin^2x$$ That means $$A=\int^{2\pi}_0\sqrt{\sin^2\frac{x}{2}}dx$$ $$A=\int^{2\pi}_0|\sin\frac{x}{2}|dx$$ As $\sin\frac{x}{2}\ge0$ on $[0,2\pi]$, $|\sin\frac{x}{2}|=\sin\frac{x}{2}$ $$A=\int^{2\pi}_0\sin\frac{x}{2}dx=2\int^{2\pi}_0\sin\frac{x}{2}d\frac{x}{2}$$ $$A=-2\cos\frac{x}{2}\Big]^{2\pi}_0$$ $$A=-2(\cos\pi-\cos0)$$ $$A=-2(-1-1)=4$$