How to evaluate the following two integral combined with anti-trigonometric function and trigonometric function？ \begin{align*}
&\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} \\
&\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx}.
\end{align*}
A few days ago,my e-friend asked me two integral,but he also doesn't know how to solve it,can somebody show me how to find their closed values?

In fact, they are similar to Coxeter’s integrals, we use Sangchul Lee's result can get 
\begin{align*}
\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} & = \frac{{{11\pi ^2}}}{72},\\
\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} & = \frac{{{\pi ^2}}}{6}.
\end{align*}
We can also try other methods!
$$\int_0^{\frac{\pi }{2}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{5 \pi ^2}{24}$$
$$\int_0^{\frac{\pi }{3}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{2 \pi ^2}{15}$$
$$\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$$
