Question

How do you find the antiderivative of `int (xsin^2x) dx`?

Answer 1

`int(xsin^2x)dx=1/4x^2-1/4xsin(2x)-1/8cos(2x)+C`

Explanation:

It's often very advantageous to rewrite `sin^2x` in integrals using the identity `cos(2x)=1-2sin^2x`. Solving for `sin^2x` in this shows that `sin^2x=1/2(1-cos(2x))`. Thus:

`intxsin^2xdx=int1/2x(1-cos(2x))dx`

Expanding:

`=1/2intxdx-1/2intxcos(2x)dx`

The first integral is basic:

`=1/2x^2/2-1/2intxcos(2x)dx`

`=1/4x^2-1/2intxcos(2x)dx`

Now performing integration by parts:

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

Note that going from `dv` to `v` requires a substitution of its own.

Then the integral equals:

`=1/4x^2-1/2(uv-intvdu)`

`=1/4x^2-1/2(1/2xsin(2x)-int1/2sin(2x)dx)`

`=1/4x^2-1/4xsin(2x)+1/4intsin(2x)dx`

Note that `intsin(2x)dx=-1/2cos(2x)`:

`=1/4x^2-1/4xsin(2x)-1/8cos(2x)+C`