# How do you find the integral of  xe^x - sec(7x)tan(7x) dx?

Apr 21, 2018

$\int \left(x {e}^{x} - \sec 7 x \tan 7 x\right) \mathrm{dx} = x {e}^{x} - {e}^{x} - \frac{1}{7} \sec \left(7 x\right) + C$

#### Explanation:

So, we want

$\int \left(x {e}^{x} - \sec 7 x \tan 7 x\right) \mathrm{dx}$. We can split up across the difference, yielding the following two integrals:

$\int x {e}^{x} \mathrm{dx} - \int \sec 7 x \tan 7 x \mathrm{dx}$

For $\int x {e}^{x} \mathrm{dx}$, we will use Integration by Parts, making the following selections:

$u = x$
$\mathrm{du} = \mathrm{dx}$
$\mathrm{dv} = {e}^{x} \mathrm{dx}$
$v = \int {e}^{x} \mathrm{dx} = {e}^{x}$

$u v - \int v \mathrm{du} = x {e}^{x} - \int {e}^{x} \mathrm{dx}$

$= x {e}^{x} - {e}^{x}$

For $\int \sec 7 x \tan 7 x \mathrm{dx}$, let's first make a simple substitution to clean things up:

$u = 7 x$
$\mathrm{du} = 7 \mathrm{dx}$
$\frac{1}{7} \mathrm{du} = \mathrm{dx}$

Then, we have the common integral

$\frac{1}{7} \int \sec u \tan u \mathrm{du} = \frac{1}{7} \sec u = \frac{1}{7} \sec \left(7 x\right)$

Combining our integrals together and putting in the constant of integration, we get

$\int \left(x {e}^{x} - \sec 7 x \tan 7 x\right) \mathrm{dx} = x {e}^{x} - {e}^{x} - \frac{1}{7} \sec \left(7 x\right) + C$