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Question-122936

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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
∫_0 ^(π/2) (dx/(b^2 tan^2 x+1)) btanx=t ⇒bsec^2 x=(dt/dx) b∫_0 ^∞ (dt/((t^2 +b^2 )(t^2 +1))) =(b/(b^2 −1))∫_0 ^∞ (1/(t^2 +1))−(1/(t^2 +b^2 )) =((πb)/(2(b^2 −1)))−(π/(2(b^2 −1))) =(π/(2(b+1)))
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:{btanx}={t}\:\:\Rightarrow{bsec}^{\mathrm{2}} {x}=\frac{{dt}}{{dx}} \\ $$$${b}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({t}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{{b}}{{b}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$=\frac{\pi{b}}{\mathrm{2}\left({b}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\pi}{\mathrm{2}\left({b}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}\left({b}+\mathrm{1}\right)} \\ $$
Commented by CanovasCamiseros last updated on 21/Nov/20
please help
$$\boldsymbol{{please}}\:\boldsymbol{{help}} \\ $$
Answered by mathmax by abdo last updated on 21/Nov/20
A =∫_0 ^(π/2) (dx/(b^2 tan^2 x +1)) ⇒ A =_(tanx=t) ∫_0 ^∞ (dt/((1+t^2 )(b^2 t^2 +1))) =_(bt=u) ∫_0 ^∞ (du/(b(1+(u^2 /b^2 ))(u^2 +1))) =∫_0 ^∞ ((bdu)/((u^2 +b^2 )(u^2 +1))) =(b/2) ∫_(−∞) ^(+∞) (du/((u^2 +b^2 )(u^2 +1))) (we rake b>0) let ϕ(z)=(1/((z^2 +b^2 )(z^2 +1))) ⇒ϕ(z)=(1/((z−ib)(z+ib)(z−i)(z+i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,ib)} Res(ϕ,i)=(1/(2i(b^2 −1))) Res(ϕ,ib) =(1/(2ib(1−b^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(2i(b^2 −1)))+(1/(2ib(1−b^2 )))} =(π/(b^2 −1)) −(π/(b(b^2 −1))) =(π/(b^2 −1))(1−(1/b)) =((π(b−1))/(b(b−1)(b+1))) =(π/(b(b+1))) ⇒ A =(b/2).(π/(b(b+1))) =(π/(2(b+1)))
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{b}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \mathrm{x}\:+\mathrm{1}}\:\Rightarrow\:\mathrm{A}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{b}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=_{\mathrm{bt}=\mathrm{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{du}}{\mathrm{b}\left(\mathrm{1}+\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\right)\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{bdu}}{\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{b}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:\left(\mathrm{we}\:\mathrm{rake}\:\mathrm{b}>\mathrm{0}\right)\:\mathrm{let} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{ib}\right)\left(\mathrm{z}+\mathrm{ib}\right)\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:+\mathrm{Res}\left(\varphi,\mathrm{ib}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\frac{\mathrm{1}}{\mathrm{2i}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{ib}\right)\:=\frac{\mathrm{1}}{\mathrm{2ib}\left(\mathrm{1}−\mathrm{b}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2i}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2ib}\left(\mathrm{1}−\mathrm{b}^{\mathrm{2}} \right)}\right\} \\ $$$$=\frac{\pi}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\:−\frac{\pi}{\mathrm{b}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{\pi}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{b}}\right)\:=\frac{\pi\left(\mathrm{b}−\mathrm{1}\right)}{\mathrm{b}\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{b}+\mathrm{1}\right)}\:=\frac{\pi}{\mathrm{b}\left(\mathrm{b}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{A}\:=\frac{\mathrm{b}}{\mathrm{2}}.\frac{\pi}{\mathrm{b}\left(\mathrm{b}+\mathrm{1}\right)}\:=\frac{\pi}{\mathrm{2}\left(\mathrm{b}+\mathrm{1}\right)} \\ $$

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