Question

How do you integrate `int x(1-x)sin(px) dx` using integration by parts?

Answer 1

`((x-x^2)cos(px))/p+((2x-1)sin(px))/p^2+(2cos(px))/p^3+C`

Explanation:

`I=int(x^2-x)sin(px)dx`

Integration takes the form: `intudv=uv-intvdu`

For this case, we will let:

`{(u=x^2-x" "=>" "du=(2x-1)dx),(dv=sin(px)dx" "=>" "v=-cos(px)/p):}`

Note that going to `dv` to `v` requires the integration of `sin(px)`, which can be done through a simple substitution.

Plugging these into the integration by parts formula, we see that:

`I=-((x^2-x)cos(px))/p-int(2x-1)(-cos(px))/pdx`

Simplifying:

`I=((x-x^2)cos(px))/p+1/p int(2x-1)cos(px)dx`

We will apply integration by parts once more for the remaining integral.

`{(u=2x-1" "=>" "du=2dx),(dv=cos(px)dx" "=>" "v=sin(px)/p):}`

Note that like before, integrating `cos(px)` is doable through substitution.

Replacing `1/p int(2x-1)cos(px)dx` with its equivalent found through integration by parts, we see that:

`I=((x-x^2)cos(px))/p+1/p[((2x-1)sin(px))/p-int(2sin(px))/pdx]`

Simplifying:

`I=((x-x^2)cos(px))/p+((2x-1)sin(px))/p^2-2/p^2intsin(px)`

This type of integration has already been performed twice before:

`I=((x-x^2)cos(px))/p+((2x-1)sin(px))/p^2-2/p^2(-cos(px)/p)`

`I=((x-x^2)cos(px))/p+((2x-1)sin(px))/p^2+(2cos(px))/p^3+C`