integral of derivated function mix $$\int_0^{2\pi}(t-\sin t){\sqrt{1-\cos t}} dt$$
I can notice that I have something of the form 
$$\int{f(x){\sqrt {f'(x)}}dx}$$ but I don't know anything that could simplify it
 A: HINT:
Note that $\sqrt{1-\cos(x)}=\sqrt{2}|\sin(x/2)|$
The integral of $x\sin(x/2)$ can be evaluated using integration by parts, while the integral of $\sin(x)\sin(x/2)$ is facilitated using the addition angle formula.
A: Well, we have that
$$\mathcal{I}:=\int_0^{2\pi}\left(t-\sin\left(t\right)\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t=$$
$$\int_0^{2\pi}t\sqrt{1-\cos\left(t\right)}\space\text{d}t-\int_0^{2\pi}\sin\left(t\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t\tag1$$
Now, we get that:


*

*Substiute $\text{u}=1-\cos\left(t\right)$:
$$\int_0^{2\pi}\sin\left(t\right)\sqrt{1-\cos\left(t\right)}\space\text{d}t=\int_0^0\sqrt{\text{u}}\space\text{d}\text{u}=0\tag2$$

*Substiute $\text{s}=\frac{t}{2}$:
$$\int_0^{2\pi}t\sqrt{1-\cos\left(t\right)}\space\text{d}t=\sqrt{2}\int_0^{2\pi}t\sin\left(\frac{t}{2}\right)\space\text{d}t=4\sqrt{2}\int_0^\pi\text{s}\sin\left(\text{s}\right)\space\text{d}\text{s}\tag3$$


Now, using integration by parts:
$$4\sqrt{2}\int_0^\pi\text{s}\sin\left(\text{s}\right)\space\text{d}\text{s}=4\sqrt{2}\cdot\left(\pi+\int_0^\pi\cos\left(\text{s}\right)\space\text{d}\text{s}\right)=$$
$$4\sqrt{2}\cdot\left(\pi+\sin\left(\pi\right)-\sin\left(0\right)\right)=4\pi\sqrt{2}\tag4$$
So, we get that:
$$\mathcal{I}=4\pi\sqrt{2}$$
A: Another approach that saves you partial integration. First remark that:
\begin{align} \int_0^{2\pi}f(t)\mathrm{d}t=\int_0^{2\pi}f(2\pi-t)\mathrm{d}t\end{align}
In this particular case you will get:
\begin{align} I=\int_0^{2\pi}(t-\sin(t))\sqrt[]{1-\cos(t)}\mathrm{d}t=\int_0^{2\pi}(2\pi-t+\sin(t))\sqrt[]{1-\cos(t)}\mathrm{d}t\end{align}
Adding them to eachother yields,
\begin{align} 2I=\int_0^{2\pi}2\pi\sqrt[]{1-\cos(t)}\mathrm{d}t\end{align}
Now use the fact:$\sqrt[]{1-\cos(t)}=\sqrt[]{2}|\sin(t/2)|$ as pointed out by @dr-mv . 
