Question Number 174430 by CElcedricjunior last updated on 31/Jul/22
Answered by som(math1967) last updated on 01/Aug/22
![I=∫_0 ^π ((xsinx)/(2(1+cos^2 x)))dx =∫_0 ^π (((π−x)sin(π−x))/(2{1+cos^2 (π−x)}))dx =(π/2)∫_0 ^π ((sinxdx)/(1+cos^2 x)) −∫_0 ^π ((xsinx)/(1+cos^2 x))dx ∴I=(π/2)∫_0 ^π ((sinxdx)/(1+cos^2 x)) −I 2I=(π/2)∫_0 ^π ((−d(cosx))/(1+cos^2 x)) 2I=(π/2)[−tan^(−1) cosx]^π _0 2I=(π/2)[−(−(π/4))+((π/4))] I=(π^2 /8)](https://www.tinkutara.com/question/Q174432.png)
$$\:{I}=\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)}{dx} \\ $$$$\:=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){sin}\left(\pi−{x}\right)}{\mathrm{2}\left\{\mathrm{1}+{cos}^{\mathrm{2}} \left(\pi−{x}\right)\right\}}{dx} \\ $$$$\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sinxdx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:−\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\therefore{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sinxdx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:−{I} \\ $$$$\:\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{−{d}\left({cosx}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}} \\ $$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\left[−\mathrm{tan}^{−\mathrm{1}} {cosx}\underset{\mathrm{0}} {\right]}^{\pi} \\ $$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\left[−\left(−\frac{\pi}{\mathrm{4}}\right)+\left(\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$\:{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$ \\ $$$$ \\ $$