What is the integral of  cos(theta)^2?

Aug 13, 2016

$= \frac{1}{2} \left[\phi + \frac{1}{2} \sin 2 \phi\right] + C$

Explanation:

Use double angle formula

$\cos 2 \phi = 2 {\cos}^{2} \phi - 1$

$\therefore {\cos}^{2} \phi = \frac{1}{2} \left(1 + \cos 2 \phi\right)$

$\int {\cos}^{2} \phi \mathrm{dp} h i = \frac{1}{2} \int \left(1 + \cos 2 \phi\right) \mathrm{dp} h i$

$= \frac{1}{2} \left[\phi + \frac{1}{2} \sin 2 \phi\right] + C$