differential equation help please...?

Question

find the general solution :

dy/dt = (2^y)*(sin t)^2

Answer 1

Separate variables:
∫(2^-y)dy = ∫((sin t)^2)dt
∫(e^(-yln(2)))dy = ∫((1/2)-cos(2t)/2)dt [as cos(x)^2-sin(x)^2 = cos(2x) and sin(x)^2 + cos(x)^2 = 1]

-e^(-ln(2)y)/ln(2) = t/2 - sin(2t)/4 + c
rearrange to get y = ....

Answer 2

dy/2^y=(sint)^2 dt
take integral of both sides

for the right hand side
take tan x/2=u
calculate dx, sinx=2sinx/2cosx/2