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Question Number 135174 by bemath last updated on 11/Mar/21

$$\mathcal{Z}={\int }_0^{\pi /2}\arctan \left(\sin \mathrm{x}\right)\mathrm{dx}+{\int }_0^{\pi /4}\arcsin \left(\tan \mathrm{x}\right)\mathrm{dx}$$

Answered by john_santu last updated on 11/Mar/21

$$let{\mathcal{Z}}_1={\int }_0^{\pi /2}\arctan \left(\sin x\right)dx$$$$settng\sin x=q\Rightarrow {\mathcal{Z}}_1={\int }_0^1\frac{\arctan \left(q\right)}{\sqrt{1-q^2}}dq$$$${\mathcal{Z}}_1={(\arctan \left(q\right).\arcsin \left(q\right)]}_0^1-{\int }_0^1\frac{\arcsin q}{1+q^2}dq$$$${\mathcal{Z}}_1=\frac{{\pi }^2}{8}-{\int }_0^1\frac{\arcsin q}{1+q^2}dq$$$$let{\mathcal{Z}}_2={\int }_0^{\pi /4}\arcsin \left(\tan x\right)dx$$$$setting\tan x=q$$$${\mathcal{Z}}_2={\int }_0^1\frac{\arcsin \left(q\right)}{1+q^2}dq$$$$Nowweget\mathcal{Z}={\mathcal{Z}}_1+{\mathcal{Z}}_2$$$$\mathcal{Z}=\frac{{\pi }^2}{8}-{\int }_0^1\frac{\arcsin \left(q\right)}{1+q^2}dq+{\int }_0^1\frac{\arcsin \left(q\right)}{1+q^2}dq$$$$\mathcal{Z}=\frac{{\pi }^2}{8}\bullet$$

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