<s>[INST] 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?
 [/INST] 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!
$$\int_0^{\frac{\pi }{2}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{5 \pi ^2}{24}$$
$$\int_0^{\frac{\pi }{3}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{2 \pi ^2}{15}$$
$$\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$$
</s>