I did the replacement, $u = \sqrt{2\sin x}$, but I did not succeed. $u = \sqrt{2\sin x}$. I found, $$ \int_{0}^{\pi/2}\arctan(x)\cot(x)\,dx, \quad \int_{0}^{\pi/2}\arctan(\sin x)\,dx $$ But the techniques used in these integrals did not help much.

Thank you for any help.

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    $\begingroup$ This leads to a hypergeometric function $$\frac{\sqrt{2 \pi } \left(9 \Gamma \left(\frac{5}{4}\right) \, _3F_2\left(\frac{1}{4},\frac{1}{4},1;\frac{3}{4},\frac{5}{4};\frac{1}{4}\right )-\frac{2 \Gamma \left(\frac{3}{4}\right)^2 \, _3F_2\left(\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};\frac{1}{4}\right )}{\Gamma \left(-\frac{3}{4}\right)}\right)}{9 \Gamma \left(\frac{3}{4}\right)}-\frac{\pi ^2}{12}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 4 at 19:00
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    $\begingroup$ It was proposed in a mathematical discussion group. The author says that the solution involves the Catalan constant. $\endgroup$ – Mathsource Mar 4 at 19:09
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    $\begingroup$ An initial step could be the use of $$\arctan(1/x)=\frac\pi2-\arctan x$$ So we have that $$I=\int_0^{\pi/2}\arctan\left[\frac1{\sqrt{2\sin x}}\right]dx=\frac{\pi^2}4-\int_0^{\pi/2}\arctan(\sqrt{2\sin x})dx$$ $\endgroup$ – clathratus Mar 4 at 19:29

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