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

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

V.G
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Sid
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    What have you tried? What is the context for your problem? – Ninad Munshi Feb 24 '21 at 03:40
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    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 '21 at 03:42
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    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. – Ninad Munshi Feb 24 '21 at 03:42
  • Look up for Coxeter’s Integrals, e.g. http://cfile22.uf.tistory.com/attach/1176F84E4F31112B111C67 – Quanto Feb 24 '21 at 04:40
  • this is a hard integral. Any answer using high-school level techniques would be very long and potentially hard to follow – clathratus Feb 24 '21 at 06:48
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    Why is this given a precalculus tag? This is like a calc. 10 integral. – BobaFret Feb 25 '21 at 04:09
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    @BobaFret It was my mistake, I didn't know that it was a tough integral, I have edited the tag. – V.G Feb 25 '21 at 04:17

1 Answers1

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Here is the method to evaluate this integral. enter image description here

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Source:- Some very challenging calculus problems by Joseph Breen

user6262
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    is there a simpler way? shorter? – Sid Feb 24 '21 at 05:45
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    I don't think that there is some shorter method. This seems long if you don't know ahmed integral.$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$ – user6262 Feb 24 '21 at 05:48