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$\int\limits^\frac{\pi}{4}_0{xcos^2x}\,dx=$

$=(\frac{2xsin2x+cos2x+2x^2}{8})|^\frac{\pi}{4}_0=$

$=(\frac{2*\frac{\pi}{4}*sin2*\frac{\pi}{4}+cos2*\frac{\pi}{4}+2*(\frac{\pi}{4})^2}{8})-(\frac{2*0*sin2*0+cos2*0+2*0^2}{8})=$

$=(\frac{\frac{\pi}{2}sin\frac{\pi}{2}+cos\frac{\pi}{2}+2*\frac{\pi^2}{16}}{8})-(\frac{0+cos0+0}{8})=$

$=\frac{\frac{\pi}{2}*1+0+\frac{\pi^2}{8}}{8}-\frac{1}{8}=$

$=\frac{\frac{\pi}{2}+\frac{\pi^2}{8}}{8}-\frac{1}{8}=\frac{\pi}{2*8}+\frac{\pi^2}{8*8}-\frac{1}{8}=$

$=\frac{\pi}{16}+\frac{\pi^2}{64}-\frac{1}{8}=-\frac{1}{8}+\frac{\pi^2}{64}+\frac{\pi}{16}$

Ответ: $-\frac{1}{8}+\frac{\pi^2}{64}+\frac{\pi}{16}$