How do you find the integral of f(x)=xnsinxn−1 using integration by parts?

1 Answer
mason m
May 3, 2018

There's no closed form.

Explanation:

I=∫xnsin(xn−1)dx

Note that ∫xn−2sin(xn−1)dx can be solved using the substitution t=xn−1, since dt=(n−1)xn−2dx and we see that the integral becomes

∫xn−2sin(xn−1)dx=1n−1∫(n−1)xn−2sin(xn−1)dx

=1n−1∫sin(t)dt=cos(xn−1)n−1

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

I=∫x2xn−2sin(xn−1)dx

Now, let dv=xn−2sin(xn−1)dx and let u=x2. As we already determined, these mean that v=cos(xn−1)n−1 and it's easy to see that du=2xdx. Then:

I=uv−∫vdu=x2cos(xn−1)n−1−2n−1∫xcos(xn−1)dx

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

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