Question

внизу pi/4 наверху pi/2 cos2xdx

Answer 1

$intlimits^frac{pi}{2}_frac{pi}{4} {cos2x} , dx =frac{1}{2}sin2x=frac{1}{2}sin(2*frac{pi}{2})-frac{1}{2}sin(2*frac{pi}{4})=frac{1}{2}sinpi-frac{1}{2}sinfrac{pi}{2}=\frac{1}{2}*0-frac{1}{2}*1=-frac{1}{2}$