Question

помогите пожалуйса!!!!

Answer 1

$intlimits^frac{pi}{4}_0{xcos^2x},dx=$

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

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

$=(frac{frac{pi}{2}sinfrac{pi}{2}+cosfrac{pi}{2}+2*frac{pi^2}{16}}{8})-(frac{0+cos0+0}{8})=$

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

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

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

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