Question Number 148856 by gsk2684 last updated on 31/Jul/21
$${pls}\:{solve} \\ $$
Commented by gsk2684 last updated on 31/Jul/21
Commented by Kamel last updated on 31/Jul/21
$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xdx}}{{sin}\left({x}\right)}\overset{{t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)} {=}\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Arctan}\left({t}\right)}{{t}}{dt}=\mathrm{2}{I}_{\mathrm{2}} \\ $$$$\therefore\:\frac{{I}_{\mathrm{1}} }{{I}_{\mathrm{2}} }=\mathrm{2}. \\ $$
Commented by gsk2684 last updated on 01/Aug/21
$${thank}\:{you} \\ $$