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السلام عليكم ورحمة الله وبركاته

\[\huge \int_{0}^{\Pi}\frac{{\color{Red} xsinx}}{{\color{Blue} 2-sin^2x}}\]

$\int_{0}^{\pi}\frac{xsinx}{2-sin^{2}x}dx$

$I=\int_{0}^{\pi}\frac{xsinx}{2-sin^{2}x}dx.......(1)$

$I=\int_{0}^{\pi}\frac{(\pi -x)sin(\pi -x)}{2-sin^{2}(\pi -x)}dx\rightarrow \int_{0}^{\pi}\frac{(\pi -x)sin(x)}{2-sin^{2}(x)}dx.....(2)$

$(1)+(2)$

$2I=\int_{0}^{\pi}\frac{\pi sinx}{2-sin^{2}x}dx\rightarrow 2I=\int_{0}^{\pi}\frac{\pi sinx}{1+cos^{2}x}dx$

$2I=-\pi arctan(cosx)]_{0}^{\pi }\rightarrow 2I=\frac{\pi ^{2}}{2}\rightarrow I=\frac{\pi ^{2}}{4}$