My attempt at $f''(x)+\sin(f'(x))=0$ does it work?

Question

Here's an attempt at a fairly innocent-looking separable differential equation. $$f''(x)+\sin f'(x)=0$$ $$\int\frac{f''(x)}{\sin f'(x)}dx=-(x+c)$$ let $y=f'(x)$ then $dy=f''(x)dx$ $$\int\csc y\,dy=-(x+c)$$ $$-\ln|\csc y+\cot y|=-(x+c)$$ $$\csc y+\cot y=e^{x+c}$$ let $a=e^{x+c}$, $w=\sin y$ $$\frac{1}{w}+\frac{\sqrt{1-w^2}}{w}=a$$ $$1+\sqrt{1-w^2}=aw$$ $$a^2w^2=1+2\sqrt{1-w^2}+1-w^2$$ $$w^2(a^2+1)=2\biggl(1+\sqrt{1-w^2}\biggr)$$ $$w^2(a^2+1)=2aw$$ $$w=\frac{2a}{a^2+1}$$ $$f'(x)=\arcsin\biggr(\frac{2a}{a^2+1}\biggl)$$ $$f(x)=\int\arcsin\biggr(\frac{2e^{x+c}}{e^{2(x+c)}+1}\biggl)dx$$ Which is a pretty nasty integral. Does this solution work? I know there is an elementary-ish form of the integral (given here) but I want to know if I can represent this with a series or hypergeometric function.