Question

What is the antiderivative of `x^2(sinx)dx`?

Answer 1

Use integration by parts, twice.

Let `u = x^2` and `dv = sinxdx`. Then `du = 2xdx` and `v = -cosx`.

`int(udv) = uv - int(vdu)`

`int(x^2sinx) = -x^2cosx - int(-cosx * 2xdx)`

`int(x^2sinx) = -x^2cosx + 2int(xcosxdx)`

Now, let `u = x` and `dv = cosxdx`. Then `du = dx` and `v = sinx`. Now, use integration by parts again.

`int(x^2sinx) = -x^2cosx + 2(xsinx - int(sinx)) + C`

`int(x^2sinx) = -x^2cosx + 2xsinx + 2cosx + C`

`int(x^2sinx) = (2 - x^2)cosx + 2xsinx + C`

Hopefully this helps!