[tex]\int\limits_0^{\pi}{x\,\sin^{2}\left(x\right)}{\;\mathrm{d}x}=\int\limits_0^{\pi}{\dfrac{x\,\left(1-\cos\left(2\,x\right)\right)}{2}}{\;\mathrm{d}x}=\dfrac{1}{2}\int\limits_0^{\pi}{\left (x-x\,\cos\left(2\,x\right) \right )}{\;\mathrm{d}x}=\\=-\dfrac{x\,\sin\left(2\,x\right)}{4}-\dfrac{\cos\left(2\,x\right)}{8}+\dfrac{{x}^{2}}{4}\Bigg|_0^{\pi}=F\left(\pi\right)=\dfrac{{\pi}^{2}}{4}-\dfrac{1}{8}-\left ( -\dfrac{1}{8} \right )=\dfrac{{\pi}^{2}}{4}[/tex]
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[tex]\int\limits_0^{\pi}{x\,\sin^{2}\left(x\right)}{\;\mathrm{d}x}=\int\limits_0^{\pi}{\dfrac{x\,\left(1-\cos\left(2\,x\right)\right)}{2}}{\;\mathrm{d}x}=\dfrac{1}{2}\int\limits_0^{\pi}{\left (x-x\,\cos\left(2\,x\right) \right )}{\;\mathrm{d}x}=\\=-\dfrac{x\,\sin\left(2\,x\right)}{4}-\dfrac{\cos\left(2\,x\right)}{8}+\dfrac{{x}^{2}}{4}\Bigg|_0^{\pi}=F\left(\pi\right)=\dfrac{{\pi}^{2}}{4}-\dfrac{1}{8}-\left ( -\dfrac{1}{8} \right )=\dfrac{{\pi}^{2}}{4}[/tex]