Question Number 123020 by mnjuly1970 last updated on 21/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}.. \\ $$$$ \\ $$$$\:\:\:{calculate}\:::\:\:\:\emptyset=\int_{\mathrm{0}} ^{\:\pi} \frac{\pi}{\mathrm{1}−{sin}\left({x}\right){cos}\left({x}\right)}{dx}=??? \\ $$$$\:\:\:\:\:\:\:\:……………….. \\ $$
Answered by Dwaipayan Shikari last updated on 21/Nov/20
![∫_0 ^π (π/(1−(1/2)sin2x))dx =2∫_0 ^∞ ((2π)/(2−((2t)/(1+t^2 )))).(1/(1+t^2 ))dt t=tanx =2∫_0 ^∞ (π/(t^2 −t+1))dt=2∫_0 ^∞ (π/((t−(1/2))+((√(3/4)))^2 ))=[((4π)/( (√3)))tan^(−1) ((2t−1)/( (√3)))]_0 ^∞ =((2π^2 )/( (√3)))](https://www.tinkutara.com/question/Q123021.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}\pi}{\mathrm{2}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:{t}={tanx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\pi}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\right)^{\mathrm{2}} }=\left[\frac{\mathrm{4}\pi}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{2}\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}} \\ $$
Commented by mnjuly1970 last updated on 21/Nov/20
$${thank}\:{you}\:{so}\:{much}… \\ $$