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\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?

Eufisky
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    This is a good question but it would be better if you ask them separately. One question, one integral only. – Tunk-Fey Aug 26 '14 at 14:10
  • I'm so sorry for that,but I think they are so similar that I want to put them together!@Tunk-Fey – Eufisky Aug 26 '14 at 14:18
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    I'll ask the usual question: do you have any reason to think that these integrals can be evaluated in closed form? Many integrals can't. – Greg Martin Aug 26 '14 at 14:46
  • Does anybody think using the definition of $\arccos x=\ln[x\pm\sqrt{x^2-1}]+\frac{\pi}{2i}+\frac{\pi}2\pm2\pi n, n=0,1,2,3,...$ would help? – Simply Beautiful Art Dec 16 '15 at 01:55

1 Answers1

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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!

$$\int_0^{\frac{\pi }{2}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{5 \pi ^2}{24}$$ $$\int_0^{\frac{\pi }{3}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{2 \pi ^2}{15}$$ $$\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$$

Eufisky
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