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

May 3, 2018

There's no closed form.

#### Explanation:

$I = \int {x}^{n} \sin \left({x}^{n - 1}\right) \mathrm{dx}$

Note that $\int {x}^{n - 2} \sin \left({x}^{n - 1}\right) \mathrm{dx}$ can be solved using the substitution $t = {x}^{n - 1}$, since $\mathrm{dt} = \left(n - 1\right) {x}^{n - 2} \mathrm{dx}$ and we see that the integral becomes

$\int {x}^{n - 2} \sin \left({x}^{n - 1}\right) \mathrm{dx} = \frac{1}{n - 1} \int \left(n - 1\right) {x}^{n - 2} \sin \left({x}^{n - 1}\right) \mathrm{dx}$

$= \frac{1}{n - 1} \int \sin \left(t\right) \mathrm{dt} = \cos \frac{{x}^{n - 1}}{n - 1}$

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

$I = \int {x}^{2} {x}^{n - 2} \sin \left({x}^{n - 1}\right) \mathrm{dx}$

Now, let $\mathrm{dv} = {x}^{n - 2} \sin \left({x}^{n - 1}\right) \mathrm{dx}$ and let $u = {x}^{2}$. As we already determined, these mean that $v = \cos \frac{{x}^{n - 1}}{n - 1}$ and it's easy to see that $\mathrm{du} = 2 x \mathrm{dx}$. Then:

$I = u v - \int v \mathrm{du} = \frac{{x}^{2} \cos \left({x}^{n - 1}\right)}{n - 1} - \frac{2}{n - 1} \int x \cos \left({x}^{n - 1}\right) \mathrm{dx}$

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