How to integrate $\int_0^{2\pi} \frac12 \sin(t) (1- \cos(t)) \sqrt{\frac12 - \frac12 \cos(t)}\,dt$ How to integrate 
$$\int_0^{2\pi} \frac12 \sin(t) (1- \cos(t)) \sqrt{\frac12 - \frac12 \cos(t)}\,dt$$
I know the solution is $0$, but I don't know how one gets this.
 A: \begin{align}
\frac{1}{2}\sin t\left(1-\cos t\right)\sqrt{\frac{1}{2}-\frac{1}{2}\cos t} &= \frac{1}{2}\sin t\left(1-\cos t\right)\sqrt{\frac{1}{2}}\sqrt{1-\cos t} \\
&= \frac{1}{2\sqrt{2}}\sin t\left(1-\cos t\right)^{3/2}\,dt
\end{align}
so let $u = 1-\cos t\to du = \sin t \,dt$ which makes the problem easier to solve.
A: Note that
$$\int_a^{b} f(t)dt=\int_a^{b} f(a+b-t)dt$$
Thus
$$I=\int_0^{2\pi} \frac12 \sin(t) (1- \cos(t)) \sqrt{\frac12 - \frac12 \cos(t)}\,dt=\int_0^{2\pi} \frac12 \sin(2\pi-t) (1- \cos(2\pi-t)) \sqrt{\frac12 - \frac12 \cos(2\pi-t)}\,dt=-\int_0^{2\pi} \frac12 \sin(t) (1- \cos(t)) \sqrt{\frac12 - \frac12 \cos(t)}\,dt=-I$$
Hence
$$2I=0$$
or
$$I=0$$
A: You can set $t=2u$, so
$$
\sqrt{\frac{1-\cos t}{2}}=\sqrt{\frac{1-1+2\sin^2u}{2}}=|\sin u|
$$
Since $0\le t\le 2\pi$ is the same as $0\le u\le \pi$, you can ignore the absolute value and your integral becomes
$$
\int_0^\pi\sin2u(1-\cos2u)\sin u\,du
=
\int_0^\pi 4\sin^4u\cos u\,du=
\left[\frac{4}{5}\sin^5u\right]_0^\pi=0
$$
But there's a much slicker way: set $t=u+\pi$, so the integral becomes
$$
-\frac{1}{2}\int_{-\pi}^{\pi}
\sin u(1+\cos u)\sqrt{\frac{1+\cos u}{2}}\,du
$$
Set $f(u)=\sin u(1+\cos u)\sqrt{\dfrac{1+\cos u}{2}}$ and note that $f(-u)=-f(u)$.
