## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{\sqrt{2}}{2}$
Use the identity $\cos 2x=2\cos^{2}x-1$ $I=\displaystyle \int_{0}^{\pi/6}\sqrt{1+\cos 2x}dx=\int_{0}^{\pi/6}\sqrt{1+(2\cos^{2}x-1)}dx$ $=\displaystyle \int_{0}^{\pi/6}\sqrt{2\cos^{2}x}dx$ $=\displaystyle \sqrt{2}\int_{0}^{\pi/6}\sqrt{\cos^{2}x}dx$ $=\displaystyle \sqrt{2}\int_{0}^{\pi/6}|\cos x|dx$ ... cosine is positive on the interval $[0,\pi/6]$ ... so we can remove the absolute value brackets $=\displaystyle \sqrt{2}\int_{0}^{\pi/6}\cos xdx$ $=\sqrt{2}[\sin x]_{0}^{\pi/6}$ $=\displaystyle \sqrt{2}(\frac{1}{2}-0)$ $=\displaystyle \frac{\sqrt{2}}{2}$