# How do you integrate int x^nsinx^(n)dx using integration by parts?

May 5, 2018

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

#### Explanation:

integration by parts

$\int {x}^{n} \sin {x}^{n} \mathrm{dx}$

suppose:

$u = \sin \left({x}^{n}\right)$

$\mathrm{du} = \cos \left({x}^{n}\right) \cdot n \cdot {x}^{n - 1} \cdot \mathrm{dx}$

$\mathrm{dv} = {x}^{n} \cdot \mathrm{dx}$

$v = \frac{1}{n + 1} \cdot {x}^{n + 1}$

$\int u \cdot \mathrm{dv} = u v - \int v \cdot \mathrm{du}$

$\int {x}^{n} \sin {x}^{n} \mathrm{dx} = \left[\frac{1}{n + 1} \cdot {x}^{n + 1}\right] \cdot \sin \left({x}^{n}\right) - \int \left[\frac{1}{n + 1} \cdot {x}^{n + 1}\right] \left[\cos \left({x}^{n}\right) \cdot n \cdot {x}^{n - 1}\right] \cdot \mathrm{dx}$

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