How to integrate tan−1(12sin⁡(x))I want to calculate the following integral :∫{0}π2tan−1(12sin⁡(x))dxBut I don't how; I tried by subsituting u=12sin⁡(x) and u=tan−1(12sin⁡(x)) but it doesn't lead me anywhere.

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How to integrate tan−1(12sin⁡(x))I want to calculate the following integral :∫{0}π2tan−1(12sin⁡(x))dxBut I don&#039;t how; I tried by subsituting u=12sin⁡(x) and u=tan−1(12sin⁡(x)) but it doesn&#039;t lead me anywhere.

tutaonana223a · Accepted Answer

Well, the integral over (0,2π)  is clearly zero by symmetry, hence the given problem is equivalent to finding−∫0π2arctan⁡(12sin⁡x)dx=−∫01arctan⁡12x1−x2dx=−π24+∫01arctan⁡(2x)1−x2dxor−π24+∫02∫01x(1+a2x2)1−x2dxda=−π24+∫02arcsinh(a)a1+a2daor−π24+∫0log⁡(2+5)usinhudu=−π24+∫12+52log⁡vv2−1dvwhere the last integral depends on the dilogarithm Li2 evaluated at ±(5−2)∫0π2arctan⁡(12sin⁡x)dx=arcsinh(t12)arcsinh(2)−Li2(5−2)+Li2(2−5)

How to integrate \(\displaystyle{{\tan}^{{-{1}}}{\left({\frac{{{1}}}{{{2}{\sin{{\left({x}\right)}}}}}}\right)}}\) I want to calculate

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