# How do you integrate int x sec 5 x dx  using integration by parts?

Feb 26, 2016

I don't believe you do.

#### Explanation:

We can start with $u = x$ and $\mathrm{dv} = \sec \left(5 x\right)$, so that

$\mathrm{du} = \mathrm{dx}$ and $v = \int \sec \left(5 x\right) \mathrm{dx} = \frac{1}{5} \ln \left(\tan 5 x + \sec 5 x\right)$

We get,

$\int x \sec \left(5 x\right) \mathrm{dx} = \frac{x}{5} \ln \left(\tan 5 x + \sec 5 x\right) - \int \ln \left(\tan 5 x + \sec 5 x\right) \mathrm{dx}$

If you have some way of finding that integral, use it.

Wolfram Alpha gives the original integral as:

$\int x \sec \left(5 x\right) \mathrm{dx} = \frac{1}{25} i \left(L {i}_{2} \left(- i {e}^{5 i x}\right) - L {i}_{2} \left(i {e}^{5 i x}\right)\right) + \frac{1}{5} x \left(\log \left(1 - i {e}^{5 i x}\right) - \log \left(1 + i {e}^{5 i x}\right)\right) + C$

Where $L {i}_{n}$ is the polylogarithmic function.