Question

Внизу pi/4 наверху pi/2 cos2xdx

Answer 1

$\int\limits^\frac{\pi}{2}_\frac{\pi}{4} {cos2x} \, dx =\frac{1}{2}sin2x=\frac{1}{2}sin(2*\frac{\pi}{2})-\frac{1}{2}sin(2*\frac{\pi}{4})=\frac{1}{2}sin\pi-\frac{1}{2}sin\frac{\pi}{2}=\\\frac{1}{2}*0-\frac{1}{2}*1=-\frac{1}{2}$