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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

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\mathrm{x}\sqrt{\mathrm{tanx}}\mathrm{dx}\mathrm{changement}\sqrt{\mathrm{tanx}}=\mathrm{t}\mathrm{give}\mathrm{tanx}={\mathrm{t}}^2\Rightarrow \mathrm{x}=\arctan \left({\mathrm{t}}^2\right)\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\mathrm{t}\arctan \left({\mathrm{t}}^2\right)\times \frac{2\mathrm{t}}{1+{\mathrm{t}}^4}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{{2\mathrm{t}}^2}{1+{\mathrm{t}}^4}\arctan \left({\mathrm{t}}^2\right)\mathrm{dt}$$$$={\int }_0^1\frac{{2\mathrm{t}}^2}{1+{\mathrm{t}}^4}\arctan \left({\mathrm{t}}^2\right)\mathrm{dt}+{\int }_1^{\mathrm{\infty }}\frac{{2\mathrm{t}}^2}{1+{\mathrm{t}}^4}\arctan \left({\mathrm{t}}^2\right)\mathrm{dt}(\to \mathrm{t}=\frac{1}{\mathrm{u}})$$$$=2{\int }_0^1\frac{{\mathrm{t}}^2\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^4}\mathrm{dt}+2{\int }_0^1\frac{1}{{\mathrm{u}}^2(1+\frac{1}{{\mathrm{u}}^4})}(\frac{\pi }{2}-\arctan \left({\mathrm{t}}^2\right))\frac{\mathrm{du}}{{\mathrm{u}}^2}$$$$=2{\int }_0^1\frac{{\mathrm{t}}^2\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^4}\mathrm{dt}+2{\int }_0^1\frac{1}{{\mathrm{u}}^4+1}(\frac{\pi }{2}-\arctan \left({\mathrm{t}}^2\right))\mathrm{dt}$$$$=2{\int }_0^1\frac{({\mathrm{t}}^2-1)\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^4}\mathrm{dt}+\pi {\int }_0^1\frac{\mathrm{du}}{{\mathrm{u}}^4+1}\mathrm{and}$$$${\int }_0^1\frac{({\mathrm{t}}^2-1)\arctan \left({\mathrm{t}}^2\right)}{1+{\mathrm{t}}^4}\mathrm{dt}={\int }_0^1({\mathrm{t}}^2-1)\arctan \left({\mathrm{t}}^2\right)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{t}}^{4\mathrm{n}}\mathrm{dt}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1({\mathrm{t}}^2-1)\arctan \left({\mathrm{t}}^2\right)\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}={\int }_0^1({\mathrm{t}}^2-1)\arctan \left({\mathrm{t}}^2\right)\mathrm{dt}...\mathrm{be}\mathrm{continued}...$$

Commented by BHOOPENDRA last updated on 19/Jan/21

$$thankyousir$$

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