# How do you integrate tan^4x sec^4x dx?

##### 1 Answer
Jul 23, 2018

$\int {\tan}^{4} x {\sec}^{4} x \mathrm{dx} = \frac{1}{7} {\tan}^{7} x + \frac{1}{5} {\tan}^{5} x + C$

#### Explanation:

When integrating a function that is a product of tangents and secants, a good strategy is to either to have

$\int f \left(\tan x\right) {\sec}^{2} x \mathrm{dx}$

or to have

$\int f \left(\sec x\right) \sec x \tan x \mathrm{dx}$.

If we can do this, we can simply integrate by substitution. This may be confusing, but looking at this concrete example will help.

$\int {\tan}^{4} x {\sec}^{4} x \mathrm{dx}$

$= \int \left({\tan}^{4} x {\sec}^{2} x\right) {\sec}^{2} x \mathrm{dx}$

By the Pythagorean identity $\textcolor{red}{{\sec}^{2} x = {\tan}^{2} x + 1}$:

$= \int \left({\tan}^{4} x \left(\textcolor{red}{{\tan}^{2} x + 1}\right)\right) {\sec}^{2} x \mathrm{dx}$

$= \int \left({\tan}^{6} x + {\tan}^{4} x\right) {\sec}^{2} x \mathrm{dx}$

Now, let $\textcolor{b l u e}{u = \tan x}$, $\textcolor{b l u e}{\mathrm{du} = {\sec}^{2} x \mathrm{dx}}$.

$= \int \left({\textcolor{b l u e}{u}}^{6} + {\textcolor{b l u e}{u}}^{4}\right) \textcolor{b l u e}{\mathrm{du}}$

$= \frac{1}{7} {u}^{7} + \frac{1}{5} {u}^{5} + C$

Undoing substitution:

$= \frac{1}{7} {\tan}^{7} x + \frac{1}{5} {\tan}^{5} x + C$