Question Number 36201 by prof Abdo imad last updated on 30/May/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\mathrm{1}+\mathrm{2}{sin}^{\mathrm{2}} \theta} \\ $$
Commented by prof Abdo imad last updated on 01/Jun/18
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{d}\theta}{\mathrm{1}+\mathrm{2}{sin}^{\mathrm{2}} \theta} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\mathrm{1}\:+\mathrm{2}\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{2}−{cos}\left(\mathrm{2}\theta\right)} \\ $$$$=_{\mathrm{2}\theta={t}} \:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\:−{cos}\left({t}\right)}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}} \:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}\left\{\mathrm{2}−\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right\}}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:−\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{{dx}}{\mathrm{1}\:+\mathrm{3}{x}^{\mathrm{2}} }\:{changement}\:\:{x}\sqrt{\mathrm{3}}\:={u}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$