Question Number 196063 by Rodier97 last updated on 17/Aug/23
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{calcul}\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\sqrt{\mathrm{u}}\:.\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by witcher3 last updated on 17/Aug/23
$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{x}\sqrt{\mathrm{tg}\left(\mathrm{x}\right)}\mathrm{dx}\:\mathrm{done} \\ $$
Commented by Rodier97 last updated on 18/Aug/23
$${how}\:?? \\ $$
Commented by Frix last updated on 18/Aug/23
$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{x}\sqrt{\mathrm{tan}\:{x}}\:{dx} \\ $$$${t}=\mathrm{tan}\:{x};\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\sqrt{{t}}\:\mathrm{tan}^{−\mathrm{1}} \:{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \:{t};\:{dt}=\left({t}^{\mathrm{2}} +\mathrm{1}\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{x}\sqrt{\mathrm{tan}\:{x}}\:{dx} \\ $$$$... \\ $$