# How do you integrate int cos6x dx?

Dec 18, 2016

$\int \cos \left(6 x\right) \mathrm{dx} = \frac{1}{6} \sin \left(6 x\right) + C$

#### Explanation:

Using integration by substitution along with the integral $\int \cos \left(x\right) \mathrm{dx} = \sin \left(x\right) + C$

let $u = 6 x \implies \mathrm{du} = 6 \mathrm{dx}$

then

$\int \cos \left(6 x\right) \mathrm{dx} = \frac{1}{6} \int \cos \left(6 x\right) 6 \mathrm{dx}$

$= \frac{1}{6} \int \cos \left(u\right) \mathrm{du}$

$= \frac{1}{6} \sin \left(u\right) + C$

$= \frac{1}{6} \sin \left(6 x\right) + C$