# Solve $\int_{0}^{\pi/2} \arccos\left( \frac{\cos(x)}{1+2\cos(x)} \right) \mathrm dx$ [closed]

$$\int_{0}^{\pi/2} \cos^{-1}\left( \dfrac{\cos(x)}{1+2\cos(x)} \right) \,dx$$

The final answer is: $$\dfrac{5\pi^2}{24}$$

• What have you tried? What is the context for your problem? Feb 24, 2021 at 3:40
• tried substituition first... there is no scope for that as far as i know... tried by parts... and it just got complicated...
– Sid
Feb 24, 2021 at 3:42
• What about your context? Why are you doing this problem? Edit all your attempts and explanations into your post, don't respond to me as a comment. Feb 24, 2021 at 3:42
• Why is this given a precalculus tag? This is like a calc. 10 integral. Feb 25, 2021 at 4:09
• @BobaFret It was my mistake, I didn't know that it was a tough integral, I have edited the tag.
– V.G
Feb 25, 2021 at 4:17

• I don't think that there is some shorter method. This seems long if you don't know ahmed integral.$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$ Feb 24, 2021 at 5:48