Question Number 27828 by abdo imad last updated on 15/Jan/18
$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}{dx}\:. \\ $$
Commented by NECx last updated on 15/Jan/18
$${thanks}\:{for}\:{this}\:{question}. \\ $$$${I}\:{really}\:{need}\:{the}\:{answer}. \\ $$
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) .](https://www.tinkutara.com/question/Q27865.png)
$${let}\:{use}\:{the}\:{ch}.\:\:\sqrt{{tanx}}={t}\:\Leftrightarrow\:\:{tanx}={t}^{\mathrm{2}} \:\:\Leftrightarrow\:{x}=\:{arctan}\:\left({t}^{\mathrm{2}} \right) \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}{dx}=\:\int_{\mathrm{0}} ^{\infty} \:{t}\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:=\int_{\mathrm{0}} ^{\propto} \:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\propto} ^{+\propto} \:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:\:\:{let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$${f}\left({z}\right)=\:\frac{\mathrm{2}{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{4}} }\:\:{let}\:{find}\:{the}\:{poles}\:{of}\:{f}? \\ $$$$\mathrm{1}+{z}^{\mathrm{4}} =\mathrm{0}\Leftrightarrow\:\:{z}^{\mathrm{4}} =\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:{so}\:{the}\:{poles}\:{are}\:{z}_{{k}\:} =\:{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \\ $$$${and}\:{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right]\:{we}\:{have}\:{z}_{\mathrm{0}} =\:{e}^{{i}\frac{\pi}{\mathrm{4}}} ,\:\:{z}_{\mathrm{1}} =\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:,\:{z}_{\mathrm{2}} =\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \\ $$$${z}_{\mathrm{3}} =\:{e}^{{i}\frac{\mathrm{7}\pi}{\mathrm{4}}} \:\:{and}\:\int_{{R}^{} } \:\:{f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\left(\:{Res}\left({f},{z}_{\mathrm{0}} \right)\:+{Res}\left({f},{z}_{\mathrm{1}} \:\right)\right) \\ $$$${Res}\left({f}\:,{z}_{\mathrm{0}} \right)=\:\:\frac{\mathrm{2}{z}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}{z}_{\mathrm{0}} ^{\mathrm{3}} }=\:\frac{\mathrm{1}}{\mathrm{2}}\:{z}_{\mathrm{0}} ^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${Res}\left({f},{z}_{\mathrm{1}} \right)=\:\frac{\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}{z}_{\mathrm{1}} ^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\:{z}_{\mathrm{1}} ^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} =\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−{i}\left(\pi−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$$\int_{{R}} \:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left[\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} −{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\right]={i}\pi\left(−\mathrm{2}{i}\:{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\mathrm{2}\pi\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\pi\sqrt{\mathrm{2}}\Rightarrow\:\:{I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} \:{f}\left({z}\right){dz}=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:. \\ $$$$ \\ $$
Answered by ajfour last updated on 15/Jan/18
$${let}\:\sqrt{\mathrm{tan}\:{x}}={t} \\ $$$$\Rightarrow\:\:\:{dx}=\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \sqrt{\mathrm{tan}\:{x}}{dx}=\int_{\mathrm{0}} ^{\:\:\infty} \:\frac{\mathrm{2}{t}^{\mathrm{2}} {dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\:\infty} \frac{\mathrm{2}{dt}}{\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} \right)} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\:\infty} \frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int_{\mathrm{0}} ^{\:\:\infty} \frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{0}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:. \\ $$