Find the particular solution to the differential equation y' = sin(x) given the initial condition y(π/2)=1

Question

Find the particular solution to the differential equation y' = sin(x) given the initial condition y(π/2)=1

Answer 1

y = -cos x+ c

1 = - cos (pi/2) + c

1 = c

so y = -cos x + 1

Answer 2

Integrating

y = -cos(x) + C

y(π/2) = 1 = -cos(π/2) + C = 0 + C --> C = 1

y = 1-cos(x)