# How do you integrate int x^3 sec^2 x dx  using integration by parts?

Mar 14, 2016

The first two steps are OK. After that you need polylogarithmic functions (beyond the scope of an introductory course).

#### Explanation:

Let $I = \int {x}^{3} {\sec}^{2} x \mathrm{dx}$

Let $u = {x}^{3}$ and $\mathrm{dv} = {\sec}^{2} x \mathrm{dx}$.

This makes $\mathrm{du} = 3 {x}^{2}$ and $v = \int {\sec}^{2} x \mathrm{dx} = \tan x$

So, we have

$u v - \int v \mathrm{du} = {x}^{3} \tan x - 3 \int {x}^{2} \tan x \mathrm{dx}$

To evaluate $\int {x}^{2} \tan x \mathrm{dx}$ use parts again.

Let $u = {x}^{2}$ and $\mathrm{dv} = \tan x \mathrm{dx}$.

So that $\mathrm{du} = 2 x \mathrm{dx}$ and $v = \int \tan x \mathrm{dx} = - \ln \left\mid \cos \right\mid x$

So we have

$I = {x}^{3} \tan x - 3 \left[- {x}^{2} \ln \left\mid \cos \right\mid x + 2 \int x \ln \left\mid \cos \right\mid x \mathrm{dx}\right]$

The last integral involves the polylogarithmic function.

I don't know enough about that to explain it. (Good luck.)