Solve the initial value problem

(1+cosx)y'=(1+e^-y)sinx y(0)=0

(1+cosx)y'=(1+e^-y)sinx y(0)=0

1/(1+e^-y) dy = sinx/(1+cosx) dx...rewrite the left side, multiplying by e^y/e^y

e^y/(e^y+1) dy = sinx/(1+cosx) dx

Ln(e^y+1)= -ln(1+cosx ) + C

At (0,0) ln(2)= -ln(2)+ C

C= 2ln2= ln4

Then ln(e^y+1) = ln4- ln(1+cosx)= ln[ 4/(1+cosx)]

e^y+1= 4/(1+cosx)

e^y= 4/(1+cosx) -1

Y= ln[ 4/(1+cosx) -1]

hoping this helps!