Solve for $y'\cos{y}=\sin(x+y)$ Solve for $y'\cos{y}=\sin(x+y)$
My attempt
$$y'\cos{y}=\sin x \cos y + \cos x \sin y$$
Divide both side by $\cos y$
$$\frac{dy}{dx}=\sin x + \cos x \tan y$$
Integrate both sides and I got
$$y = -\cos x + \int \cos x \tan y \,dx$$
As Maximilian Janisch said that $\int\cos() \tan(())d≠\sin() \tan(())$ I didn't know what to do next.

As said in comments, power series solution seems to be the only way.
Writing
$$y=\sum_{n=0}^\infty \frac{a_n}{n!} x^n$$ the very first terms would be
$$\left(
\begin{array}{cc}
 1 & t \\
 2 & t^3+t+1 \\
 3 & 3 t^5+4 t^3+t^2+1 \\
 4 & 15 t^7+27 t^5+7 t^4+9 t^3+8 t^2-3 t \\
 5 & 105 t^9+240 t^7+63 t^6+144 t^5+103 t^4+6 t^3+34 t^2-2 t-6 \\
 6 & 945 t^{11}+2625 t^9+693 t^8+2250 t^7+1476 t^6+566 t^5+847 t^4+49 t^3+48 t^2+53
   t-15
\end{array}
\right)$$ where $t=\tan(a_0)$.
Not very funny, isn't it ?
