Question

How do you find the integral of `f(x)=x^3sinx^2` using integration by parts?

Answer 1

See the explanation section below.

Explanation:

Sometimes our first choice of `u` and `dv` is not helpful. In that case, go back and try choosing differently.

If we choose `u = sinx^2` and `dv = x^3 dx`, then

`du = 2xcosx^2 dx` and `v = 1/4 x^4`.

Unfortunately we get `intvdu = 1/2int x^5cosx^2dx`.

We have made the problem worse.

Go back to the beginning. If we choose to integrate `sinx^2`, we'll need to use substitution. To use substitution, we'll need the derivative of `x^2`. We cna insert the `2`, but where can we get the `x` we need?

AH!, yes

`int x^3 sinx^2dx = intunderbrace(x^2)_u underbrace(sinx^2 * x dx)_(dv)`

Let `u = x^2` and `dv = sinx^2 * x dx`, which gets us

`du = 2xdx` and `v = -1/2cosx^2`

`uv-intvdu = -1/2x^2cosx^2 + int cosx^2 x dx`

The integral here is like the integration to find `v` above.

Students can finish this for themselves.