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

.......advanced ... .... ... calculus..... I:=∫_((−π)/2) ^( (π/2)) sin^2 (tan(x))dx=^(???) (π/e)sinh(1)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:.......{advanced}\:...\:....\:...\:{calculus}..... \\ $$$$\:\:\:\boldsymbol{\mathrm{I}}:=\int_{\frac{−\pi}{\mathrm{2}}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \left({tan}\left({x}\right)\right){dx}\overset{???} {=}\frac{\pi}{{e}}{sinh}\left(\mathrm{1}\right) \\ $$

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

∫_(−∞) ^∞ ((sin^2 (t))/(t^2 +1))dt=τ(1) Knowing ∫_(−∞) ^∞ ((cos(αx))/(x^2 +1))=πe^(−α) τ(α)=(1/2)∫_(−∞) ^∞ ((1−cos(2αx))/(x^2 +1))dx =(1/2)(π−πe^(−2α) ) =(π/2)(((e^(2α) −1)/e^(2α) )) τ(1)=(π/2)(((e^2 −1)/e^2 ))=(π/2)(((e−(1/e))/e)) =(π/e)sinh(1)

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

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

∫_(−π/2) ^(π/2) sin^2 (tan x)dx=∫_(−∞) ^(+∞) ((sin^2 u)/(u^2 +1))du=∫_0 ^∞ ((1−cos 2u)/(u^2 +1))du =(π/2)−∫_0 ^∞ ((cos 2u)/(u^2 +1))du=(π/2)−ℜ∫_0 ^∞ (e^(i2u) /(u^2 +1))du =(π/2)−ℜ{πiRes((e^(i2u) /(u^2 +1)),i)} =(π/2)−(π/(2e^2 ))

$$\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{tan}\:{x}\right){dx}=\int_{−\infty} ^{+\infty} \frac{\mathrm{sin}\:^{\mathrm{2}} {u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\frac{\pi}{\mathrm{2}}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\frac{\pi}{\mathrm{2}}−\Re\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{i}\mathrm{2}{u}} }{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\frac{\pi}{\mathrm{2}}−\Re\left\{\pi{iRes}\left(\frac{{e}^{{i}\mathrm{2}{u}} }{{u}^{\mathrm{2}} +\mathrm{1}},{i}\right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$

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

I=∫_(−(π/2)) ^(π/2) sin^2 (tanx)dx ⇒I=_(tanx=t) ∫_(−∞) ^(+∞) ((sin^2 (t))/(1+t^2 ))dt =∫_(−∞) ^(+∞) ((1−cos(2t))/(2(t^2 +1)))dt =(1/2)∫_(−∞) ^(+∞) (dt/(t^2 +1))−(1/2)∫_(−∞) ^(+∞) ((cos(2t))/(t^2 +1))dt =(π/2)−(1/2)Re(∫_(−∞) ^(+∞) (e^(2it) /(t^2 +1))dt) let w(z)=(e^(2iz) /(z^2 +1)) (=(e^(2iz) /((z−i)(z+i)))) ∫_(−∞) ^(+∞) w(z)dz=2iπ.Res(w,i) =2iπ.(e^(−2) /(2i))=(π/e^2 ) ⇒ I=(π/2)−(π/(2e^2 )) ⇒I=(π/2)(1−(1/e^2 ))

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

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