Question Number 123023 by mnjuly1970 last updated on 21/Nov/20
$$\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$${calculate}::: \\ $$$$\:\:\:\:\:\mathrm{I}:\overset{???} {=}\:\int_{\mathrm{0}} ^{\:\pi} \frac{{x}}{\mathrm{1}−{sin}\left({x}\right){cos}\left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………………….. \\ $$
Answered by mnjuly1970 last updated on 24/Nov/20
![evaluation: I=^? ∫_0 ^( π) (x/(1−sin(x)cos(x)))dx solution: I=[∫_0 ^(π/2) (x/(1−sin(x)cos(x)))dx=I_1 ]+[∫_((π/2) ) ^( π) (x/(1−xsin(x)cos(x)))dx=I_2 ] I_1 =∫_0 ^(π/2) ((2x)/(2−sin(2x)))dx=π∫_0 ^(π/2) (1/(2−sin(2x)))dx−2∫_0 ^(π/2) (x/(2−sin(2x)))dx 2I_1 =π∫_0 ^( (π/2)) (dx/(2−((2tan(x))/(1+tan^2 (x)))))=π∫_0 ^(π/2) ((1+tan^2 (x))/(2tan^2 (x)−2tan(x)+2))dx =^(tan(x)=y) (π/2)∫_0 ^∞ (dy/(y^2 −y+1))=(π/2) (2/( (√3)))[tan^(−1) ((2/( (√3)))(y−(1/2)))]_0 ^∞ =(π/( (√3)))[(π/2)+(π/6)]=((2π^2 )/( 3(√3))) ⇒ I_1 =(π^2 /(3(√3))) ✓ I_2 =∫_(π/2) ^( π) (x/(1−sin(x)cos(x)))dx =((3π)/2)∫_(π/2) ^( π) (2/(2−sin(2x)))dx−I_2 2I_2 =_I_1 ^(similarly) (((3π)/2)( (2/( (√3))))[tan^(−1) ((2/( (√3)))(t−(1/2))]_(−∞) ^0 2I_2 =((3π)/( (√3)))[((−π)/6)+(π/2)]=((3π^2 )/(3(√3))) ⇒ I_2 =(π^2 /( 2(√3))) ∴ I=I_1 +I_2 =(π^2 /( (√3)))((1/2)+(1/3))=((5π^2 )/(6(√3))) ✓✓ r](https://www.tinkutara.com/question/Q123241.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{evaluation}:\:\:\mathrm{I}\overset{?} {=}\int_{\mathrm{0}} ^{\:\pi} \frac{{x}}{\mathrm{1}−{sin}\left({x}\right){cos}\left({x}\right)}{dx} \\ $$$$\:\:\:{solution}: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{I}=\left[\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{\mathrm{1}−{sin}\left({x}\right){cos}\left({x}\right)}{dx}=\mathrm{I}_{\mathrm{1}} \right]+\left[\int_{\frac{\pi}{\mathrm{2}}\:} ^{\:\pi} \frac{{x}}{\mathrm{1}−{xsin}\left({x}\right){cos}\left({x}\right)}{dx}=\mathrm{I}_{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\mathrm{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{x}}{\mathrm{2}−{sin}\left(\mathrm{2}{x}\right)}{dx}=\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}−{sin}\left(\mathrm{2}{x}\right)}{dx}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{\mathrm{2}−{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$\:\:\:\:\:\mathrm{2I}_{\mathrm{1}} =\pi\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{2}−\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}}=\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{tan}^{\mathrm{2}} \left({x}\right)−\mathrm{2}{tan}\left({x}\right)+\mathrm{2}}{dx} \\ $$$$\:\:\overset{{tan}\left({x}\right)={y}} {=}\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dy}}{{y}^{\mathrm{2}} −{y}+\mathrm{1}}=\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\overset{\infty} {\right]}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}\left[\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right]=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\:\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\:\mathrm{I}_{\mathrm{1}} =\frac{\pi^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\:\checkmark\: \\ $$$$\:\:\:\:\:\mathrm{I}_{\mathrm{2}} =\int_{\frac{\pi}{\mathrm{2}}} ^{\:\pi} \frac{{x}}{\mathrm{1}−{sin}\left({x}\right){cos}\left({x}\right)}{dx} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{2}}} ^{\:\pi} \frac{\mathrm{2}}{\mathrm{2}−{sin}\left(\mathrm{2}{x}\right)}{dx}−\mathrm{I}_{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{2I}_{\mathrm{2}} \underset{\mathrm{I}_{\mathrm{1}} } {\overset{{similarly}} {=}}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\left(\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right]_{−\infty} ^{\mathrm{0}} \right.\right. \\ $$$$\:\:\:\:\:\mathrm{2I}_{\mathrm{2}} =\frac{\mathrm{3}\pi}{\:\sqrt{\mathrm{3}}}\left[\frac{−\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{2}}\right]=\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\:\mathrm{I}_{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\:\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\therefore\:\mathrm{I}=\mathrm{I}_{\mathrm{1}} +\mathrm{I}_{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{6}\sqrt{\mathrm{3}}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$${r} \\ $$