Question

How do you integrate `int x^2sinx` by integration by parts method?

Answer 1

`intx^2sinx dx = 2cosx + 2 xsinx -x^2cos x+ C`

Explanation:

The formula for integration by parts is:
`intu(dv)/dxdx = uv - intv(du)/dxdx`

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

In this case a trig function stays as a trig function but `x^2` simplifies if differentiated.

Let `{(u=x^2, => ,(du)/dx=2x),((dv)/dx=sinx,=>,v =-cosx ):}`

So IBP gives:

`int(x^2)(sinx)dx = (x^2)(-cos x)-int(-cosx)(2x)dx`
`:. intx^2sinx dx = -x^2cos x + 2intxcosxdx` ..... [1]

Ok, well that is better, but for the second interval we have to use IBP again;

Let `{(u=x, => ,(du)/dx=1),((dv)/dx=cosx,=>,v =sinx ):}`

So IBP gives:

`int (x)(cosx)dx = (x)(sinx) - int (sinx)(1)dx`
`:. int xcosxdx = xsinx - int sinxdx`
`:. int xcosxdx = xsinx + cosx` ..... [2]

Substituting [2] into [1] we have;

`intx^2sinx dx = -x^2cos x + 2{ xsinx + cosx} + C`
`:. intx^2sinx dx = -x^2cos x + 2 xsinx + 2cosx + C`
`:. intx^2sinx dx = 2cosx + 2 xsinx -x^2cos x+ C`