Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 253: 56

$y=\frac{-cos(2z)}{2}+4$

$\frac{dy}{dz}=sin(2z);$ $y(\frac{\pi}{4})=4$ $dy=sin(2z)dz$ $\int dy=\int sin(2z)dz$ $y+C=\frac{1}{2}\int2sin(2z)dz$ $y=\frac{1}{2}(-cos(2z))+C$ Substitute in initial condition to solve for $C$ $4=\frac{-cos(\frac{\pi}{2})}{2}+C$ $4=C$ $y=\frac{-cos(2z)}{2}+4$