Evaluate : `intx^(2)sinxdx`.

Explanation

Integrating by parts, taking `x^(2)` as the first function, we get
`intx^(2)sinxdx=x^(2)intsinxdx-int[(d)/(dx)(x^(2))*intsinxdx]dx`
`=x^(2)(-cosx)-int2x(-cosx)dx`
`=-x^(2)cosx+2intxcosxdx`
`=-x^(2)cosx+2[x(sinx)-int{(d)/(dx)(x)*intcosxdx}dx]`
[integrating x cos x by parts]
`=-x^(2)cosx+2[xsinx-intsinxdx]`
`=-x^(2)cosx+2[xsinx+cosx]+C`.