Question Number 190841 by Rupesh123 last updated on 12/Apr/23
Answered by ARUNG_Brandon_MBU last updated on 12/Apr/23
![I=∫_0 ^π ((xdx)/(1+cosαsinx))=∫_0 ^π ((π−x)/(1+cosαsinx))dx =(1/2)∫_0 ^π (π/(1+cosαsinx))dx=(1/2)∫_0 ^∞ (π/(1+cosα((2t)/(1+t^2 ))))∙((2dt)/(1+t^2 )) =π∫_0 ^∞ (dt/(t^2 +2tcosα+1))=π∫_0 ^∞ (dt/((t+cosα)^2 +sin^2 α)) =(π/(sinα))[arctan(((t+cosα)/(sinα)))]_0 ^∞ =(π/(sinα))((π/2)−arctan((1/(tanα)))) =(π/(sinα))(arctan(tanα))=((πα)/(sinα))](Q190843.png)
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{xdx}}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\mathrm{1}+\mathrm{cos}\alpha\mathrm{sin}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\mathrm{1}+\mathrm{cos}\alpha\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\:\:\:=\pi\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}{t}\mathrm{cos}\alpha+\mathrm{1}}=\pi\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({t}+\mathrm{cos}\alpha\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \alpha} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{sin}\alpha}\left[\mathrm{arctan}\left(\frac{{t}+\mathrm{cos}\alpha}{\mathrm{sin}\alpha}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{sin}\alpha}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{tan}\alpha}\right)\right) \\ $$$$\:\:\:=\frac{\pi}{\mathrm{sin}\alpha}\left(\mathrm{arctan}\left(\mathrm{tan}\alpha\right)\right)=\frac{\pi\alpha}{\mathrm{sin}\alpha} \\ $$
Commented by Rupesh123 last updated on 12/Apr/23
Nice solution, sir!
Commented by mehdee42 last updated on 13/Apr/23
$${very}\:{good} \\ $$