Question

How do you find the integral of `f(x)=x^nsinx^(n-1)` using integration by parts?

Answer 1

There's no closed form.

Explanation:

`I=intx^nsin(x^(n-1))dx`

Note that `intx^(n-2)sin(x^(n-1))dx` can be solved using the substitution `t=x^(n-1)`, since `dt=(n-1)x^(n-2)dx` and we see that the integral becomes

`intx^(n-2)sin(x^(n-1))dx =1/(n-1)int(n-1)x^(n-2)sin(x^(n-1))dx`

`=1/(n-1)intsin(t)dt=cos(x^(n-1))/(n-1)`

Motivated by this integrable function, let's rewrite `I`:

`I=intx^2x^(n-2)sin(x^(n-1))dx`

Now, let `dv=x^(n-2)sin(x^(n-1))dx` and let `u=x^2`. As we already determined, these mean that `v=cos(x^(n-1))/(n-1)` and it's easy to see that `du=2xdx`. Then:

`I=uv-intvdu=(x^2cos(x^(n-1)))/(n-1)-2/(n-1)intxcos(x^(n-1))dx`

Unfortunately, this integral has no closed form. Did you type your question right?