# How do you integrate int sec^4(5x)?

Jan 14, 2017

$\tan \frac{5 x}{5} + {\tan}^{3} \frac{5 x}{15} + C$

#### Explanation:

Start first with a simple substitution to simplify the argument of the secant functions: let $u = 5 x$ which implies that $\mathrm{du} = 5 \mathrm{dx}$. Then:

$\int {\sec}^{4} \left(5 x\right) \mathrm{dx} = \frac{1}{5} \int {\sec}^{4} \left(5 x\right) \left(5 \mathrm{dx}\right) = \frac{1}{5} \int {\sec}^{4} \left(u\right) \mathrm{du}$

When working with secant, it's important to keep the identities $\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$, $\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$, and $1 + {\tan}^{2} x = {\sec}^{2} x$ in mind.

In this case, we see that the integral

$= \frac{1}{5} \int {\sec}^{2} \left(u\right) {\sec}^{2} \left(u\right) \mathrm{du} = \frac{1}{5} \int \left(1 + {\tan}^{2} \left(u\right)\right) {\sec}^{2} \left(u\right) \mathrm{du}$

The point of doing this is that we have a function of tangent—$1 + {\tan}^{2} \left(u\right)$—paired with the derivative of tangent—${\sec}^{2} \left(u\right)$.

So, let $v = \tan \left(u\right)$ so that $\mathrm{dv} = {\sec}^{2} \left(u\right) \mathrm{du}$. This gives us:

$= \frac{1}{5} \int \left(1 + {v}^{2}\right) \mathrm{dv}$

We can now integrate term by term:

$= \frac{1}{5} \left(v + {v}^{3} / 3\right) = \frac{v}{5} + {v}^{3} / 15$

Returning to our original variable $x$ using $v = \tan \left(u\right)$ and $u = 5 x$ then adding the constant of integration:

$= \tan \frac{u}{5} + {\tan}^{3} \frac{u}{15} = \tan \frac{5 x}{5} + {\tan}^{3} \frac{5 x}{15} + C$