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

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

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    $\begingroup$ What have you tried? What is the context for your problem? $\endgroup$ Feb 24, 2021 at 3:40
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    $\begingroup$ tried substituition first... there is no scope for that as far as i know... tried by parts... and it just got complicated... $\endgroup$
    – Sid
    Feb 24, 2021 at 3:42
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    $\begingroup$ 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. $\endgroup$ Feb 24, 2021 at 3:42
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    $\begingroup$ Why is this given a precalculus tag? This is like a calc. 10 integral. $\endgroup$
    – BobaFret
    Feb 25, 2021 at 4:09
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    $\begingroup$ @BobaFret It was my mistake, I didn't know that it was a tough integral, I have edited the tag. $\endgroup$
    – V.G
    Feb 25, 2021 at 4:17

1 Answer 1

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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

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    $\begingroup$ is there a simpler way? shorter? $\endgroup$
    – Sid
    Feb 24, 2021 at 5:45
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    $\begingroup$ I don't think that there is some shorter method. This seems long if you don't know ahmed integral.$\int_{0}^{1}\frac{\tan^{-1}\sqrt{x^{2}+2}}{(x^{2}+1)\sqrt{x^{2}+2}}\mathop{\mathrm{d}x}=\frac{5\pi ^{2}}{96}$ $\endgroup$
    – user6262
    Feb 24, 2021 at 5:48

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