While solving problem 9 of Brilliant Integration Contest Season 1, I did the following -

\[\begin{align}I= \int_{0}^{1} \log \left(\dfrac{1+x}{1-x}\right) \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x &= 2 \int_{0}^{1}\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1} \dfrac{1}{x\sqrt{1-x^2}} \ \mathrm{d}x\\ &=2\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2n}}{(2n+1)\sqrt{1-x^2}} \ \mathrm{d}x \end{align}\] Then I used the substitution \(x \mapsto \sin \theta \).

\[\begin{align} \therefore I &=2\sum_{n=0}^{\infty} \int_{0}^{\pi/2} \dfrac{\sin^{2n} {\theta}}{2n+1} \ \mathrm{d}\theta\\ &=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)} \end{align}\] The last step is due to Wallis' formula.

**Now, my question is that how do we evaluate this last series?**

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TopNewestHave to changed the limits to obtain the consequences in using wallis formula.

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I didn't get what you are saying.. could you explain?

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You are in class 12 and you know such a high level calculus , can you give any suggestions or a recommened a book for increasing my ability in calculas@Pratik Shastri

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Well, @Krishna Sharma just posted a similar question as a problem. here. It is not exactly the same but quite a bit yeah. And it has a solution too.

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I posted this long ago :P I now know how to get the result.

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Can you post the result here ?

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