## Thomas' Calculus 13th Edition

$\frac{x}{2}+\frac{\sin2x}{4}+C$
Recall: $\cos 2x= 2\cos^{2}x-1$ or $\cos^{2}x= \frac{1+\cos2x}{2}$ We have, $\int \cos^{2}xdx= \frac{1}{2}\int (1+\cos2x)dx$ $=\frac{1}{2}\int dx+\frac{1}{2}\int \cos 2xdx$ $=\frac{x}{2}+\frac{\sin2x}{4}+C$