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Question Number 136651 by mhabs last updated on 24/Mar/21

Answered by Ñï= last updated on 24/Mar/21

$${\int }_0^{\pi /2}\frac{1}{1+{\left(a\tan x\right)}^2}dx\stackrel{t=\tan x}{=}{\int }_0^{\mathrm{\infty }}\frac{1}{1+a^2{t}^2}\cdot \frac{1}{1+t^2}dt$$$$=\frac{1}{1-a^2}{\int }_0^{\mathrm{\infty }}(\frac{-a^2}{1+a^2{t}^2}+\frac{1}{1+t^2})dt=\frac{1}{1-a^2}\{\frac{\pi }{2}-\frac{a^2}{a}{\tan }^{-1}\left(at\right){\mid }_0^{\mathrm{\infty }}\}$$$$=\frac{1}{1-a^2}(1-a)\frac{\pi }{2}$$$$=\frac{\pi }{2(1+a)}$$

Answered by Dwaipayan Shikari last updated on 24/Mar/21

$${\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\left(atan\left(x\right)\right)}^2}dxatan\left(x\right)=u{asec}^2\left(x\right)=\frac{du}{dx}$$$$=a{\int }_0^{\mathrm{\infty }}\frac{1}{(1+u^2)(a^2+u^2)}du=\frac{a}{a^2-1}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}-\frac{1}{a^2+u^2}du$$$$=\frac{\pi a}{2(a^2-1)}-\frac{\pi }{2}=\frac{\pi }{2(a+1)}$$

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