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Question Number 129679 by BHOOPENDRA last updated on 17/Jan/21

Answered by mathmax by abdo last updated on 18/Jan/21

I=∫_0 ^(π/2) x(√(tanx))dx changement (√(tanx))=t give tanx=t^2 ⇒x=arctan(t^2 ) ⇒ I=∫_0 ^∞ t arctan(t^2 )×((2t)/(1+t^4 ))dt =∫_0 ^∞ ((2t^2 )/(1+t^4 )) arctan(t^2 )dt =∫_0 ^1 ((2t^2 )/(1+t^4 ))arctan(t^2 )dt +∫_1 ^∞ ((2t^2 )/(1+t^4 )) arctan(t^2 )dt(→t=(1/u)) =2∫_0 ^1 ((t^2 arctan(t^2 ))/(1+t^4 ))dt +2∫_0 ^1 (1/(u^2 (1+(1/u^4 ))))((π/2)−arctan(t^2 ))(du/u^2 ) =2∫_0 ^1 ((t^2 arctan(t^2 ))/(1+t^4 ))dt +2∫_0 ^1 (1/(u^4 +1))((π/2)−arctan(t^2 ))dt =2∫_0 ^1 (((t^2 −1)arctan(t^2 ))/(1+t^4 ))dt +π∫_0 ^1 (du/(u^4 +1)) and ∫_0 ^1 (((t^2 −1)arctan(t^2 ))/(1+t^4 ))dt =∫_0 ^1 (t^2 −1)arctan(t^2 )Σ_(n=0) ^∞ (−1)^n t^(4n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 (t^2 −1)arctan(t^2 )dt =Σ_(n=0) ^∞ (−1)^n u_n u_n =∫_0 ^1 (t^2 −1)arctan(t^2 )dt...be continued...

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{x}\sqrt{\mathrm{tanx}}\mathrm{dx}\:\:\mathrm{changement}\:\sqrt{\mathrm{tanx}}=\mathrm{t}\:\mathrm{give}\:\mathrm{tanx}=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}=\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{t}\:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)×\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:+\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\left(\rightarrow\mathrm{t}=\frac{\mathrm{1}}{\mathrm{u}}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{4}} }\right)}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right)\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{4}} \:+\mathrm{1}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\right)\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:+\pi\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{4}} \:+\mathrm{1}}\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{t}^{\mathrm{4n}} \mathrm{dt} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\mathrm{dt}...\mathrm{be}\:\mathrm{continued}... \\ $$

Commented by BHOOPENDRA last updated on 19/Jan/21

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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