# How do you integrate int sec^6(3x)?

Jan 10, 2017

$\tan \left(3 x\right) + \frac{2}{3} {\tan}^{3} \left(3 x\right) + \frac{1}{5} {\tan}^{5} \left(3 x\right) + C$

#### Explanation:

First let $u = 3 x$ so that $\mathrm{du} = 3 \mathrm{dx}$. Then we have:

$I = \int {\sec}^{6} \left(3 x\right) \mathrm{dx} = \frac{1}{3} \int {\sec}^{6} \left(3 x\right) 3 \mathrm{dx} = \frac{1}{3} \int {\sec}^{6} \left(u\right) \mathrm{du}$

We will now make use of the identity ${\tan}^{2} \left(\theta\right) + 1 = {\sec}^{2} \left(\theta\right)$.

$I = \frac{1}{3} \int {\sec}^{4} \left(u\right) {\sec}^{2} \left(u\right) \mathrm{du} = \frac{1}{3} \int {\left({\sec}^{2} \left(u\right)\right)}^{2} {\sec}^{2} \left(u\right) \mathrm{du}$

$I = \frac{1}{3} \int {\left(1 + {\tan}^{2} \left(u\right)\right)}^{2} {\sec}^{2} \left(u\right) \mathrm{du}$

Now we should perform the substitution $v = \tan \left(u\right)$. Note that $\mathrm{dv} = {\sec}^{2} \left(u\right) \mathrm{du}$, which is why we left a ${\sec}^{2} \left(u\right)$ floating around in the integrand.

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

Expand ${\left(1 + {v}^{2}\right)}^{2} = \left(1 + {v}^{2}\right) \left(1 + {v}^{2}\right)$ and then integrate term by term:

$I = \int \left(1 + 2 {v}^{2} + {v}^{4}\right) \mathrm{dv} = v + \frac{2}{3} {v}^{3} + \frac{1}{5} {v}^{5}$

From $v = \tan \left(u\right)$:

$I = \tan \left(u\right) + \frac{2}{3} {\tan}^{3} \left(u\right) + \frac{1}{5} {\tan}^{5} \left(u\right)$

From $u = 3 x$:

$I = \tan \left(3 x\right) + \frac{2}{3} {\tan}^{3} \left(3 x\right) + \frac{1}{5} {\tan}^{5} \left(3 x\right) + C$