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Question Number 147060 by ArielVyny last updated on 17/Jul/21
∫_0 ^(π/2) e^(2x) (√(tanx))dx
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{\mathrm{2}{x}} \sqrt{{tanx}}{dx} \\ $$
Answered by mathmax by abdo last updated on 17/Jul/21
Ψ=∫_0 ^(π/(2 )) e^(2x) (√(tanx))dx changement (√(tanx))=z give x=arctan(z^2 ) Ψ=∫_0 ^∞ e^(2arctan(z^2 )) z ×((2z)/(1+z^4 ))dz =2∫_0 ^∞ ((z^2 e^(2arctan(z^2 )) )/(1+z^4 ))dz ϕ(z)=((z^2 e^(2arctan(z^2 )) )/(z^4 +1)) ⇒ϕ(z)=((z^2 e^(2arctan(z^2 )) )/((z^2 −i)(z^2 +i))) =((z^2 e^(2arctan(z^2 )) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) we can use rdsidus but this way is not sure...! ∫_R ϕ(z)dz=2iπ{Res(ϕ ,e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =((ie^(2arctan(i)) )/(2e^((iπ)/4) (2i)))=(1/4)e^(−((iπ)/4)) e^(2arctan(i)) Res(ϕ,−e^(−((iπ)/4)) ) =(((−i)e^(2arctan(−i)) )/(−2e^(−((iπ)/4)) (−2i)))=−(1/4)e^((iπ)/4) e^(2arctan(−i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz=((iπ)/2){e^(−((iπ)/4)) .e^(2arctan(i)) +e^((iπ)/4) e^(2arctan(−i)) } ...be continued....
$$\Psi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:\mathrm{e}^{\mathrm{2x}} \sqrt{\mathrm{tanx}}\mathrm{dx}\:\:\mathrm{changement}\:\sqrt{\mathrm{tanx}}=\mathrm{z}\:\mathrm{give}\:\mathrm{x}=\mathrm{arctan}\left(\mathrm{z}^{\mathrm{2}} \right) \\ $$$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{z}^{\mathrm{2}} \right)} \mathrm{z}\:×\frac{\mathrm{2z}}{\mathrm{1}+\mathrm{z}^{\mathrm{4}} }\mathrm{dz}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{z}^{\mathrm{2}} \right)} }{\mathrm{1}+\mathrm{z}^{\mathrm{4}} }\mathrm{dz} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{z}^{\mathrm{2}} \right)} }{\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{z}^{\mathrm{2}} \right)} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{i}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{z}^{\mathrm{2}} \right)} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)}\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{rdsidus} \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{way}\:\mathrm{is}\:\mathrm{not}\:\mathrm{sure}…! \\ $$$$\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\varphi\:,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{ie}^{\mathrm{2arctan}\left(\mathrm{i}\right)} }{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2i}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \mathrm{e}^{\mathrm{2arctan}\left(\mathrm{i}\right)} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\left(−\mathrm{i}\right)\mathrm{e}^{\mathrm{2arctan}\left(−\mathrm{i}\right)} }{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{2i}\right)}=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{\mathrm{2arctan}\left(−\mathrm{i}\right)} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\frac{\mathrm{i}\pi}{\mathrm{2}}\left\{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} .\mathrm{e}^{\mathrm{2arctan}\left(\mathrm{i}\right)} \:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{\mathrm{2arctan}\left(−\mathrm{i}\right)} \right\} \\ $$$$…\mathrm{be}\:\mathrm{continued}…. \\ $$

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