# How do you integrate intsec3x?

Feb 8, 2017

$\int \sec \left(3 x\right) \mathrm{dx} = \frac{1}{3} \ln \left\mid \sec \left(3 x\right) + \tan \left(3 x\right) \right\mid + C$

#### Explanation:

$\int \sec \left(3 x\right) \mathrm{dx}$

Use the substitution $u = 3 x$, implying that $\mathrm{du} = \left(3\right) \mathrm{dx}$. Then:

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

This is a common integral: $\int \sec \left(u\right) \mathrm{du} = \ln \left\mid \sec \left(u\right) + \tan \left(u\right) \right\mid + C$

We can derive this integral by multiplying the integrand by $\frac{\sec \left(u\right) + \tan \left(u\right)}{\sec \left(u\right) + \tan \left(u\right)}$:

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

Now let $v = \sec \left(u\right) + \tan \left(u\right)$. This implies that $\mathrm{dv} = \left(\sec \left(u\right) \tan \left(u\right) + {\sec}^{2} \left(u\right)\right) \mathrm{du}$. This is the numerator:

$= \frac{1}{3} \int \frac{\mathrm{dv}}{v} = \frac{1}{3} \ln \left\mid v \right\mid + C = \frac{1}{3} \ln \left\mid \sec \left(u\right) + \tan \left(u\right) \right\mid + C$

Finally:

$= \frac{1}{3} \ln \left\mid \sec \left(3 x\right) + \tan \left(3 x\right) \right\mid + C$