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Question Number 122936 by CanovasCamiseros last updated on 21/Nov/20

Commented by Dwaipayan Shikari last updated on 21/Nov/20

$${\int }_0^{\frac{\pi }{2}}\frac{dx}{b^2{tan}^{2}x+1}btanx=t\Rightarrow {bsec}^{2}x=\frac{dt}{dx}$$$$b{\int }_0^{\mathrm{\infty }}\frac{dt}{(t^2+b^2)(t^2+1)}$$$$=\frac{b}{b^2-1}{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+1}-\frac{1}{t^2+b^2}$$$$=\frac{\pi b}{2(b^2-1)}-\frac{\pi }{2(b^2-1)}$$$$=\frac{\pi }{2(b+1)}$$

Commented by CanovasCamiseros last updated on 21/Nov/20

$$\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$

Answered by mathmax by abdo last updated on 21/Nov/20

$$\mathrm{A}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{\mathrm{b}}^2{\tan }^2\mathrm{x}+1}\Rightarrow \mathrm{A}{=}_{\mathrm{tanx}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{(1+{\mathrm{t}}^2)({\mathrm{b}}^2{\mathrm{t}}^2+1)}$$$${=}_{\mathrm{bt}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{du}}{\mathrm{b}(1+\frac{{\mathrm{u}}^2}{{\mathrm{b}}^2})({\mathrm{u}}^2+1)}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{bdu}}{({\mathrm{u}}^2+{\mathrm{b}}^2)({\mathrm{u}}^2+1)}$$$$=\frac{\mathrm{b}}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{du}}{({\mathrm{u}}^2+{\mathrm{b}}^2)({\mathrm{u}}^2+1)}(\mathrm{we}\mathrm{rake}\mathrm{b}>0)\mathrm{let}$$$$\phi \left(\mathrm{z}\right)=\frac{1}{({\mathrm{z}}^2+{\mathrm{b}}^2)({\mathrm{z}}^2+1)}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{1}{(\mathrm{z}-\mathrm{ib})(\mathrm{z}+\mathrm{ib})(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\mathrm{i})+\mathrm{Res}(\phi ,\mathrm{ib})\}$$$$\mathrm{Res}(\phi ,\mathrm{i})=\frac{1}{2\mathrm{i}({\mathrm{b}}^2-1)}$$$$\mathrm{Res}(\phi ,\mathrm{ib})=\frac{1}{2\mathrm{ib}(1-{\mathrm{b}}^2)}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\frac{1}{2\mathrm{i}({\mathrm{b}}^2-1)}+\frac{1}{2\mathrm{ib}(1-{\mathrm{b}}^2)}\}$$$$=\frac{\pi }{{\mathrm{b}}^2-1}-\frac{\pi }{\mathrm{b}({\mathrm{b}}^2-1)}=\frac{\pi }{{\mathrm{b}}^2-1}(1-\frac{1}{\mathrm{b}})=\frac{\pi (\mathrm{b}-1)}{\mathrm{b}(\mathrm{b}-1)(\mathrm{b}+1)}=\frac{\pi }{\mathrm{b}(\mathrm{b}+1)}\Rightarrow$$$$\mathrm{A}=\frac{\mathrm{b}}{2}.\frac{\pi }{\mathrm{b}(\mathrm{b}+1)}=\frac{\pi }{2(\mathrm{b}+1)}$$

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