Integral of $\arccos$ within $\sin$ How to calculate the following integral:
$\int^{R}_{0}[2 \cos^{-1}(\frac{r}{2R}) -\sin(2 \cos^{-1}(\frac{r}{2R}) )  ] dr$.
This is a part of a complex formula.
 A: You can use substitution to solve this problem. Clearly the substitution is $u=\cos^{-1}\left(\frac{r}{2R}\right)$ and hence $\cos u=\frac{r}{2R}$. Then $r=0\to u=\frac{\pi}{2},r=R\to u=\frac{\pi}{3}$ and $dr=-2R\sin u$. Thus
\begin{eqnarray*}
I&=&-2R\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}(2u-\sin(2u))\sin udu\\
&=&2R\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(2u\sin u-\sin(2u)\sin u)du
\end{eqnarray*}
Now
\begin{eqnarray*}
\int(2u\sin u-\sin(2u)\sin u)du&=&-2\int ud\cos u-2\int\sin^2u\cos udu\\
&=&-2\int ud\cos u-2\int\sin^2ud\sin u\\
&=&-2u\cos u-\frac{2}{3}\sin^3u+2\int\cos udu\\
&=&-2u\cos u-\frac{2}{3}\sin^3u+2\sin u+C.
\end{eqnarray*}
So
\begin{eqnarray*}
I&=&\left.2R(-2u\cos u-\frac{2}{3}\sin^3u+2\sin u)\right|_{\frac{\pi}{3}}^{\frac{\pi}{2}}
&=&2R\cdot\frac{1}{12}(16-9\sqrt{3}+4\pi)\\
&=&\frac{R}{6}(16-9\sqrt{3}+4\pi).
\end{eqnarray*}
A: Hints:
$$\left(\sin(\arccos x)\right)'=-\cos(\arccos x)\frac{1}{\sqrt{1-x^2}}=-\frac{x}{\sqrt{1-x^2}}=\left(\sqrt{1-x^2}\right)'\implies$$
$$\sin(\arccos x)=\sqrt{1-x^2}+K\;,\;\;K=\text{ constant}$$
Choosing $\,x=1\,$ we find that $\,\sin 0=0+K\,\implies K=0$, so using now
$$\sin2\alpha=2\sin\alpha\cos\alpha\implies \sin\left(2\arccos\frac{r}{2R}\right)=2\sqrt{1-\frac{r^2}{4R^2}}\cdot\frac{r}{2R}$$
so you have
$$\int\limits_0^R\left(2\arccos\frac{r}{2R}-\frac{r}{4R^2}\sqrt{4R^2-r^2}\right)dr$$
Finally, integrating by parts:
$$\int\arccos x\,dx=x\arccos x-\sqrt{1-x^2}+C\ldots$$
