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Question Number 87534 by mathmax by abdo last updated on 04/Apr/20

$$calculate{\int }_0^{\frac{\pi }{4}}\frac{arctan\left(sinx\right)}{sinx}dx$$

Commented by mathmax by abdo last updated on 06/Apr/20

$$letf\left(a\right)={\int }_0^{\frac{\pi }{4}}\frac{arctan\left(asinx\right)}{sinx}dxwitha>0$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\frac{\pi }{4}}\frac{sinx}{(1+a^2{sin}^{2}x)sinx}dx={\int }_0^{\frac{\pi }{4}}\frac{dx}{1+a^2\times \frac{1-cos\left(2x\right)}{2}}$$$$=2{\int }_0^{\frac{\pi }{4}}\frac{dx}{2+a^2-a^{2}cos\left(2x\right)}{=}_{2x=t}2{\int }_0^{\frac{\pi }{2}}\frac{dt}{2(2+a^2-a^{2}cost)}$$$${=}_{tan\left(\frac{t}{2}\right)=u}{\int }_0^1\frac{2du}{(1+u^2)(2+a^2-a^2\frac{1-u^2}{1+u^2})}$$$$={\int }_0^1\frac{2du}{2+a^2+(2+a^2)u^2-a^2+a^2{u}^2}={\int }_0^1\frac{2du}{2+(2+2a^2)u^2}$$$$={\int }_0^1\frac{du}{1+(1+a^2)u^2}{=}_{\sqrt{1+a^2}u=z}{\int }_0^{\sqrt{1+a^2}}\frac{dz}{\sqrt{1+a^2}(1+z^2)}$$$$=\frac{1}{\sqrt{1+a^2}}arctan\left(\sqrt{1+a^2}\right)\Rightarrow f\left(a\right)={\int }_0^a\frac{arctan\left(\sqrt{1+{\alpha }^2}\right)}{\sqrt{1+{\alpha }^2}}d\alpha +c$$$$c=f\left(0\right)=0\Rightarrow f\left(a\right)={\int }_0^a\frac{arctan(\sqrt{1+{\alpha }^2)}}{\sqrt{1+{\alpha }^2}}d\alpha$$$$and{\int }_0^{\frac{\pi }{4}}\frac{arctan\left(sinx\right)}{sinx}dx={\int }_0^1\frac{arctan\left(\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}dx....becontinued...$$

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