How do you find the antiderivative of  cos^4 (x) dx?

Jun 28, 2016

You want to split it up using trig identities to get nice, easy integrals.

Explanation:

${\cos}^{4} \left(x\right) = {\cos}^{2} \left(x\right) \cdot {\cos}^{2} \left(x\right)$

We can deal with the ${\cos}^{2} \left(x\right)$ easily enough by rearranging the double angle cosine formula.

${\cos}^{4} \left(x\right) = \frac{1}{2} \left(1 + \cos \left(2 x\right)\right) \cdot \frac{1}{2} \left(1 + \cos \left(2 x\right)\right)$

${\cos}^{4} \left(x\right) = \frac{1}{4} \left(1 + 2 \cos \left(2 x\right) + {\cos}^{2} \left(2 x\right)\right)$

${\cos}^{4} \left(x\right) = \frac{1}{4} \left(1 + 2 \cos \left(2 x\right) + \frac{1}{2} \left(1 + \cos \left(4 x\right)\right)\right)$

${\cos}^{4} \left(x\right) = \frac{3}{8} + \frac{1}{2} \cdot \cos \left(2 x\right) + \frac{1}{8} \cdot \cos \left(4 x\right)$

So,

$\int {\cos}^{4} \left(x\right) \mathrm{dx} = \frac{3}{8} \cdot \int \mathrm{dx} + \frac{1}{2} \cdot \int \cos \left(2 x\right) \mathrm{dx} + \frac{1}{8} \cdot \int \cos \left(4 x\right) \mathrm{dx}$

$\int {\cos}^{4} \left(x\right) \mathrm{dx} = \frac{3}{8} x + \frac{1}{4} \cdot \sin \left(2 x\right) + \frac{1}{32} \cdot \sin \left(4 x\right) + C$