# How do I evaluate the following integral by parts? intx^9cos(x^5)dx

## $\int {x}^{9} \cos \left({x}^{5}\right) \mathrm{dx}$

Jan 18, 2018

$I = \frac{{x}^{5} \sin \left({x}^{5}\right) + \cos \left({x}^{5}\right)}{5} + C$

#### Explanation:

We seek:

$I = \int \setminus {x}^{9} \setminus \cos \left({x}^{5}\right) \setminus \mathrm{dx}$

We should first reduce the term ${x}^{5}$ in the integral by performing a substitution:

Let $\setminus t = {x}^{5} \implies \frac{\mathrm{dt}}{\mathrm{dx}} = 5 {x}^{4}$

So we can manipulate the integral, and perform this substitution to get

$I = \int \setminus \frac{1}{5} \setminus {x}^{5} \setminus \cos \left({x}^{5}\right) \setminus \left(5 {x}^{4}\right) \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{5} \setminus \int \setminus t \setminus \cos t \setminus \mathrm{du}$

And now we can apply Integration By Parts:

Let $\left\{\begin{matrix}u & = t & \implies \frac{\mathrm{du}}{\mathrm{dt}} & = 1 \\ \frac{\mathrm{dv}}{\mathrm{dt}} & = \cos t & \implies v & = \sin t\end{matrix}\right.$

Then plugging into the IBP formula:

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

We have:

$\int \setminus t \setminus \cos t \setminus \mathrm{dt} = \left(t\right) \left(\sin t\right) - \int \setminus \left(\sin t\right) \left(1\right) \setminus \mathrm{dt}$
$\text{ } = t \sin t + \cos t \setminus \setminus \left(+ c\right)$

Allowing us to write:

$I = \frac{1}{5} \left\{t \sin t + \cos t\right\} + C$

Then restoring the substitution, we get:

$I = \frac{{x}^{5} \sin \left({x}^{5}\right) + \cos \left({x}^{5}\right)}{5} + C$