Question

Вычислить интеграл Римана

Answer 1

$\int\limits^ \pi _0 {(x^2sin^2x)} \, dx =1/2 \int\limits^ \pi _0 {x^2(1-cos2x)} \, dx =$$1/2 \int\limits^ \pi _0 {(x^2-x^2cos2x)} \, dx =-1/2 \int\limits^ \pi _0 {(x^2cos2x-x^2)} \, dx =$
u=x²,du=2xdx,dv=cos2xdx,v=1/2sin2x
=$-1/4x^2sin2x+1/2 \int\limits {xsin2x} \, dx +1/2 \int\limits {x^2} \, dx=$
u=x,du=dx,dv=sin2xdx,v=1/2cos2x
=$-1/4x^2sin2x-1/4xcos2x+1/4 \int\limits {cos2x} \, dx +1/2 \int\limits {x^2} \, dx =$$-1/4x^2sin2x-1/4xcosx+1/8sin2x+x^3/6|^ \pi _0=$$- \pi ^2/4*sin2 \pi - \pi /4*cos2 \pi +1/8*sin2 \pi + \pi ^3/6+0+0-1/8*sin0$$-0=0- \pi /4+0+ \pi ^3/6=- \pi /4+ \pi ^3/6$