∫_0 ^(π/2) e^(2x) (√(tanx))dx


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Question Number 147060 by ArielVyny last updated on 17/Jul/21

$\int_{0}^{\frac{π}{2}} e^{2 x} \sqrt{t a n x} d x$
	Answered by mathmax by abdo last updated on 17/Jul/21
	$Ψ=∫_0 ^(π/(2 )) e^(2x) (√(tanx))dx changement (√(tanx))=z give x=arctan(z^2 ) Ψ=∫_0 ^∞ e^(2arctan(z^2 )) z ×((2z)/(1+z^4 ))dz =2∫_0 ^∞ ((z^2 e^(2arctan(z^2 )) )/(1+z^4 ))dz ϕ(z)=((z^2 e^(2arctan(z^2 )) )/(z^4 +1)) ⇒ϕ(z)=((z^2 e^(2arctan(z^2 )) )/((z^2 −i)(z^2 +i))) =((z^2 e^(2arctan(z^2 )) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) we can use rdsidus but this way is not sure...! ∫_R ϕ(z)dz=2iπ{Res(ϕ ,e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =((ie^(2arctan(i)) )/(2e^((iπ)/4) (2i)))=(1/4)e^(−((iπ)/4)) e^(2arctan(i)) Res(ϕ,−e^(−((iπ)/4)) ) =(((−i)e^(2arctan(−i)) )/(−2e^(−((iπ)/4)) (−2i)))=−(1/4)e^((iπ)/4) e^(2arctan(−i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz=((iπ)/2){e^(−((iπ)/4)) .e^(2arctan(i)) +e^((iπ)/4) e^(2arctan(−i)) } ...be continued....$
	$Ψ = \int_{0}^{\frac{π}{2}} e^{2 x} \sqrt{tanx} dx changement \sqrt{tanx} = z give x = \arctan (z^{2})$ $Ψ = \int_{0}^{\infty} e^{2 \arctan (z^{2})} z \times \frac{2 z}{1 + z^{4}} dz = 2 \int_{0}^{\infty} \frac{z^{2} e^{2 \arctan (z^{2})}}{1 + z^{4}} dz$ $φ (z) = \frac{z^{2} e^{2 \arctan (z^{2})}}{z^{4} + 1} \Rightarrow φ (z) = \frac{z^{2} e^{2 \arctan (z^{2})}}{(z^{2} - i) (z^{2} + i)}$ $= \frac{z^{2} e^{2 \arctan (z^{2})}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} we can use rdsidus$ $but this way is not sure . . .!$ $\int_{R} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})}$ $Res (φ, e^{\frac{i π}{4}}) = \frac{{ie}^{2 \arctan (i)}}{{2 e}^{\frac{i π}{4}} (2 i)} = \frac{1}{4} e^{- \frac{i π}{4}} e^{2 \arctan (i)}$ $Res (φ, - e^{- \frac{i π}{4}}) = \frac{(- i) e^{2 \arctan (- i)}}{- {2 e}^{- \frac{i π}{4}} (- 2 i)} = - \frac{1}{4} e^{\frac{i π}{4}} e^{2 \arctan (- i)}$ $\Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = \frac{i π}{2} {e^{- \frac{i π}{4}} . e^{2 \arctan (i)} + e^{\frac{i π}{4}} e^{2 \arctan (- i)}}$ $. . . be continued . . . .$
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