Question Number 26507 by yesaditya22@gmail.com last updated on 26/Dec/17
$$\overset{\Pi} {\int}_{\mathrm{0}} \frac{\mathrm{xsinx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$
Answered by kaivan.ahmadi last updated on 26/Dec/17
![I=∫_0 ^π ((xsinx)/(2−sin^2 x))dx. set f(x)=(x/(2−x^2 ))⇒ f(sinx)=f(sin(π−x))⇒ ∫_0 ^π ((xsinx)/(2−sin^2 x))dx=(π/2)∫_0 ^π f(sinx)dx= (π/2)∫_0 ^π ((sinx)/(1+cos^2 x)) now if u=cosx⇒du=−sinxdx I=−(π/2)∫_1 ^(−1) (du/(1+u^2 ))=−(π/2)Arctgu]_1 ^(−1) = −(π/2)(arctg(−1)−arctg1)= −(π/2)(−(π/4)−(π/4))=(π^2 /4) ⇒](https://www.tinkutara.com/question/Q26542.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsinx}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}.\:\:\mathrm{set}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{sinx}\right)=\mathrm{f}\left(\mathrm{sin}\left(\pi−\mathrm{x}\right)\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsinx}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{sinx}\right)\mathrm{dx}= \\ $$$$\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sinx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\mathrm{now}\:\mathrm{if}\:\mathrm{u}=\mathrm{cosx}\Rightarrow\mathrm{du}=−\mathrm{sinxdx} \\ $$$$\left.\mathrm{I}=−\frac{\pi}{\mathrm{2}}\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }=−\frac{\pi}{\mathrm{2}}\mathrm{Arctgu}\right]_{\mathrm{1}} ^{−\mathrm{1}} = \\ $$$$−\frac{\pi}{\mathrm{2}}\left(\mathrm{arctg}\left(−\mathrm{1}\right)−\mathrm{arctg1}\right)= \\ $$$$−\frac{\pi}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow \\ $$