Answer
$-\frac{\sqrt 2}{4}+\frac{1}{2}$
Work Step by Step
$\int_{0}^{\pi/8} (sin(2x)) dx=\frac{1}{2}\cdot\int_{0}^{\pi/8} (2sin(2x)) dx=[-\frac{1}{2}\cdot cos(2x)]_{0}^{\pi/8}=[-\frac{1}{2}\cdot cos(2\cdot\frac{\pi}{8})]-[-\frac{1}{2}\cdot cos(2\cdot0)]=[-\frac{1}{2}\cdot cos(\frac{\pi}{4})]-[-\frac{1}{2}\cdot cos(0)]=[-\frac{1}{2}\cdot(\frac{\sqrt 2}{2})]+\frac{1}{2}\cdot1=-\frac{\sqrt 2}{4}+\frac{1}{2}$
