Answer
The solution of the given initial-value problem
$$
1+\cos x) y^{\prime}=\left(1+e^{-y}\right) \sin x, \quad\quad y(0)=y
$$
is
$$
y(x)=\ln \left[4 e^{-\ln (1+\cos x)}-1\right] =\ln \frac{3-\cos x}{1+\cos x} .
$$
Work Step by Step
The initial-value problem is
$$
1+\cos x) y^{\prime}=\left(1+e^{-y}\right) \sin x, \quad\quad y(0)=y
$$
We write the equation in terms of differentials and integrate both sides:
$$
(1+\cos x) y^{\prime}=\left(1+e^{-y}\right) \sin x \Rightarrow \frac{d y}{1+e^{-y}}=\frac{\sin x d x}{1+\cos x}
$$
$ \Rightarrow$
$$
\int \frac{d y}{1+1 / e^{y}}=\int \frac{\sin x d x}{1+\cos x}
$$
$ \Rightarrow$
$$
\int \frac{e^{y} d y}{1+e^{y}}=\int \frac{\sin x d x}{1+\cos x}
$$
$$
\ln \left|1+e^{y}\right|=-\ln |1+\cos x|+C \Rightarrow \ln \left(1+e^{y}\right)=-\ln (1+\cos x)+C
$$
where $ C $ is an arbitrary constant.
$$
1+e^{y}=e^{-\ln (1+\cos x)} \cdot e^{C} \Rightarrow e^{y}=k e^{-\ln (1+\cos x)}-1
$$
where $ k=e^{C} $ is an arbitrary constant.
$ \Rightarrow$
$$
y=\ln \left[k e^{-\ln (1+\cos x)}-1\right].
$$
Since $y(0)=0$
$$
0=\ln \left[k e^{-\ln 2}-1\right] \Rightarrow e^{0}=k\left(\frac{1}{2}\right)-1 \Rightarrow k=4
$$
Thus the solution of the given initial-value problem is
$$
y(x)=\ln \left[4 e^{-\ln (1+\cos x)}-1\right] =\ln \frac{3-\cos x}{1+\cos x}
$$
