# How do you integrate int cos^3(x/3)dx?

Jan 9, 2017

$3 \sin \left(\frac{x}{3}\right) - {\sin}^{3} \left(\frac{x}{3}\right) + C$

#### Explanation:

First let $t = \frac{x}{3}$. This implies that $\mathrm{dt} = \frac{1}{3} \mathrm{dx}$. We then see that:

$\int {\cos}^{3} \left(\frac{x}{3}\right) \mathrm{dx} = 3 \int {\cos}^{3} \left(\frac{x}{3}\right) \frac{1}{3} \mathrm{dx} = 3 \int {\cos}^{3} \left(t\right) \mathrm{dt}$

To do this, split up ${\cos}^{3} \left(t\right)$ into ${\cos}^{2} \left(t\right) \cos \left(t\right)$ and then rewrite ${\cos}^{2} \left(t\right)$ using the identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1 \implies {\cos}^{2} \left(\theta\right) = 1 - {\sin}^{2} \left(\theta\right)$.

$3 \int {\cos}^{3} \left(t\right) \mathrm{dt} = 3 \int {\cos}^{2} \left(t\right) \cos \left(t\right) \mathrm{dt} = 3 \int \left(1 - {\sin}^{2} \left(t\right)\right) \cos \left(t\right) \mathrm{dt}$

Now let $s = \sin \left(t\right)$, so $\mathrm{ds} = \cos \left(t\right) \mathrm{dt}$. Luckily we already have this in the integrand!

$3 \int \left(1 - {\sin}^{2} \left(t\right)\right) \cos \left(t\right) \mathrm{dt} = 3 \int \left(1 - {s}^{2}\right) \mathrm{ds}$

Integrating term by term using $\int {s}^{n} \mathrm{ds} = {s}^{n + 1} / \left(n + 1\right)$ where $n \ne - 1$.

$3 \int \left(1 - {s}^{2}\right) \mathrm{ds} = 3 \left(s - {s}^{3} / 3\right) = 3 s - {s}^{3}$

From $s = \sin \left(t\right)$ and $t = \frac{x}{3}$ we see that $s = \sin \left(\frac{x}{3}\right)$. Also add the constant of integration.

$3 s - {s}^{3} = 3 \sin \left(\frac{x}{3}\right) - {\sin}^{3} \left(\frac{x}{3}\right) + C$