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Question Number 138524 by mnjuly1970 last updated on 14/Apr/21

$$.......advanced..........calculus.....$$$$\mathbf{I}:={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{sin}^2\left(tan\left(x\right)\right)dx\stackrel{???}{=}\frac{\pi }{e}sinh\left(1\right)$$

Answered by Dwaipayan Shikari last updated on 14/Apr/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{sin}^2\left(t\right)}{t^2+1}dt=\tau \left(1\right)$$$$Knowing{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{x^2+1}=\pi e^{-\alpha }$$$$\tau \left(\alpha \right)=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1-cos\left(2\alpha x\right)}{x^2+1}dx$$$$=\frac{1}{2}(\pi -\pi e^{-2\alpha })=\frac{\pi }{2}\left(\frac{e^{2\alpha }-1}{e^{2\alpha }}\right)$$$$\tau \left(1\right)=\frac{\pi }{2}\left(\frac{e^2-1}{e^2}\right)=\frac{\pi }{2}\left(\frac{e-\frac{1}{e}}{e}\right)$$$$=\frac{\pi }{e}sinh\left(1\right)$$

Answered by Ñï= last updated on 14/Apr/21

$${\int }_{-\pi /2}^{\pi /2}\sin {}^2\left(\tan x\right)dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sin {}^{2}u}{u^2+1}du={\int }_0^{\mathrm{\infty }}\frac{1-\cos 2u}{u^2+1}du$$$$=\frac{\pi }{2}-{\int }_0^{\mathrm{\infty }}\frac{\cos 2u}{u^2+1}du=\frac{\pi }{2}-\mathrm{\Re }{\int }_0^{\mathrm{\infty }}\frac{e^{i2u}}{u^2+1}du$$$$=\frac{\pi }{2}-\mathrm{\Re }\left\{\pi iRes(\frac{e^{i2u}}{u^2+1},i)\right\}$$$$=\frac{\pi }{2}-\frac{\pi }{2e^2}$$

Answered by mathmax by abdo last updated on 15/Apr/21

$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^2\left(\mathrm{tanx}\right)\mathrm{dx}\Rightarrow \mathrm{I}{=}_{\mathrm{tanx}=\mathrm{t}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\sin }^2\left(\mathrm{t}\right)}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1-\cos \left(2\mathrm{t}\right)}{2({\mathrm{t}}^2+1)}\mathrm{dt}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}-\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos \left(2\mathrm{t}\right)}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\frac{\pi }{2}-\frac{1}{2}\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{2\mathrm{it}}}{{\mathrm{t}}^2+1}\mathrm{dt}\right)\mathrm{let}\mathrm{w}\left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{{\mathrm{z}}^2+1}(=\frac{{\mathrm{e}}^{2\mathrm{iz}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})})$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi .\mathrm{Res}(\mathrm{w},\mathrm{i})=2\mathrm{i}\pi .\frac{{\mathrm{e}}^{-2}}{2\mathrm{i}}=\frac{\pi }{{\mathrm{e}}^2}\Rightarrow$$$$\mathrm{I}=\frac{\pi }{2}-\frac{\pi }{{2\mathrm{e}}^2}\Rightarrow \mathrm{I}=\frac{\pi }{2}(1-\frac{1}{{\mathrm{e}}^2})$$

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