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Question Number 27828 by abdo imad last updated on 15/Jan/18

find the value of   ∫_0 ^(π/2) (√(tanx))dx .

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}{dx}\:. \\ $$

Commented by NECx last updated on 15/Jan/18

thanks for this question.  I really need the answer.

$${thanks}\:{for}\:{this}\:{question}. \\ $$$${I}\:{really}\:{need}\:{the}\:{answer}. \\ $$

Commented by abdo imad last updated on 15/Jan/18

let use the ch.  (√(tanx))=t ⇔  tanx=t^2   ⇔ x= arctan (t^2 )  I= ∫_0 ^(π/2) (√(tanx))dx= ∫_0 ^∞  t ((2t)/(1+t^4 )) dt  =∫_0 ^∝    ((2t^2 )/(1+t^4 ))dt  = (1/2) ∫_(−∝) ^(+∝)    ((2t^2 )/(1+t^4 ))dt   let introduce the complex function  f(z)= ((2z^2 )/(1+z^4 ))  let find the poles of f?  1+z^4 =0⇔  z^4 = e^(i(2k+1)π)  so the poles are z_(k ) = e^(i(2k+1)(π/4))   and k∈[[0,3]] we have z_0 = e^(i(π/4)) ,  z_1 = e^(i((3π)/4))   , z_2 = e^(i((5π)/4))   z_3 = e^(i((7π)/4))   and ∫_R^    f(z)dz= 2iπ( Res(f,z_0 ) +Res(f,z_1  ))  Res(f ,z_0 )=  ((2z_0 ^2 )/(4z_0 ^3 ))= (1/2) z_0 ^(−1) =(1/2) e^(−i(π/4))   Res(f,z_1 )= ((2z_1 ^2 )/(4z_1 ^3 ))=(1/2) z_1 ^(−1) =−(1/2) e^(−i((3π)/4)) =(1/2) e^(−i(π−(π/4)))   = −(1/2) e^(i(π/4))   ∫_R  f(z)dz=2iπ[ (1/2)( e^(−i(π/4)) −e^(i(π/4)) )]=iπ(−2i sin((π/4)))  =2π ((√2)/2) =π(√2)⇒  I= (1/2) ∫_R  f(z)dz= ((π(√2))/2)  .

$${let}\:{use}\:{the}\:{ch}.\:\:\sqrt{{tanx}}={t}\:\Leftrightarrow\:\:{tanx}={t}^{\mathrm{2}} \:\:\Leftrightarrow\:{x}=\:{arctan}\:\left({t}^{\mathrm{2}} \right) \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}{dx}=\:\int_{\mathrm{0}} ^{\infty} \:{t}\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:=\int_{\mathrm{0}} ^{\propto} \:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\propto} ^{+\propto} \:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:\:{let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$${f}\left({z}\right)=\:\frac{\mathrm{2}{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{4}} }\:\:{let}\:{find}\:{the}\:{poles}\:{of}\:{f}? \\ $$$$\mathrm{1}+{z}^{\mathrm{4}} =\mathrm{0}\Leftrightarrow\:\:{z}^{\mathrm{4}} =\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{poles}\:{are}\:{z}_{{k}\:} =\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \\ $$$${and}\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]\:{we}\:{have}\:{z}_{\mathrm{0}} =\:{e}^{{i}\frac{\pi}{\mathrm{4}}} ,\:\:{z}_{\mathrm{1}} =\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:,\:{z}_{\mathrm{2}} =\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \\ $$$${z}_{\mathrm{3}} =\:{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:\:{and}\:\int_{{R}^{} } \:\:{f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\left(\:{Res}\left({f},{z}_{\mathrm{0}} \right)\:+{Res}\left({f},{z}_{\mathrm{1}} \:\right)\right) \\ $$$${Res}\left({f}\:,{z}_{\mathrm{0}} \right)=\:\:\frac{\mathrm{2}{z}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}{z}_{\mathrm{0}} ^{\mathrm{3}} }=\:\frac{\mathrm{1}}{\mathrm{2}}\:{z}_{\mathrm{0}} ^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${Res}\left({f},{z}_{\mathrm{1}} \right)=\:\frac{\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}{z}_{\mathrm{1}} ^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\:{z}_{\mathrm{1}} ^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\left(\pi−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{{R}} \:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left[\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} −{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right]={i}\pi\left(−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\mathrm{2}\pi\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\pi\sqrt{\mathrm{2}}\Rightarrow\:\:{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} \:{f}\left({z}\right){dz}=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:. \\ $$$$ \\ $$

Answered by ajfour last updated on 15/Jan/18

let (√(tan x))=t  ⇒   dx=((2tdt)/(1+t^4 ))  ∫_0 ^(  π/2) (√(tan x))dx=∫_0 ^(  ∞)  ((2t^2 dt)/(1+t^4 ))       =∫_0 ^(  ∞) ((2dt)/(((1/t^2 )+t^2 )))    =∫_0 ^(  ∞) (((1−(1/t^2 ))dt)/((t+(1/t))^2 −2))+∫_0 ^(  ∞) (((1+(1/t^2 ))dt)/((t−(1/t))^2 +2))  =(1/(2(√2))) ln (((t+(1/t)−(√2))/(t+(1/t)+(√2))))∣_0 ^∞ +(1/(√2))tan^(−1) (((t−(1/t))/(√2)))∣_0 ^∞   =0+(1/(√2))((π/2)+(π/2)) =(π/(√2)) .

$${let}\:\sqrt{\mathrm{tan}\:{x}}={t} \\ $$$$\Rightarrow\:\:\:{dx}=\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \sqrt{\mathrm{tan}\:{x}}{dx}=\int_{\mathrm{0}} ^{\:\:\infty} \:\frac{\mathrm{2}{t}^{\mathrm{2}} {dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\:\infty} \frac{\mathrm{2}{dt}}{\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} \right)} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\:\infty} \frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int_{\mathrm{0}} ^{\:\:\infty} \frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{0}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\sqrt{\mathrm{2}}}\:. \\ $$

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