How do you solve $y'=\sin(y)$? 
How do you solve $y'=\sin(y)$?

I ended up with
$$-\ln(\csc(y)+\cot(y)) = x+C$$
But then I simplified it to
$$e^{-x+C} = \frac1{\sin(y)} + \frac{\cos(y)}{\sin(y)}$$
Where do I go from here?
 A: It is better to write
$$-\log(\csc y+\cot y)=\log\tan\frac y2$$
Then we have
$$\log\tan\frac y2=x+K$$
$$\tan\frac y2=Ae^x$$
$$y=2\tan^{-1}Ae^x$$
A: It's better to use the Weierstrass substitution :
$$I=\int \dfrac {dy}{\sin y}=\int \dfrac {1+t^2}{2t}\dfrac {2}{1+t^2}dt$$
$$I=\int \dfrac {dt}{t}=\ln t +C= \ln \left|\tan \dfrac y 2 \right|+C$$
Where $t=\tan (y/2)$
A: $$y'=\sin y\iff\frac{dy}{\sin y}=dx\iff \int\frac{dy}{\sin y}=\int dx\implies$$
$$\implies\ln\tan\frac y2+c\implies$$
$$\implies\tan\frac y2=e^{c+x}\implies \frac y2=\arctan\left(e^ce^x\right)\implies\ldots$$
A: Notice that for any integer $k$, the constant function $y(x)=\pi k$ is a solution because
$$\sin\left(y(x)\right)=\sin(\pi k)=0=y'(x)$$
To find the others, we may assume that $y(x)$ is not of the form $\pi k$. Then $\sin(y(x))$ is not identically zero, so
$$\csc(y(x))y'(x)=1$$
Antiderivatives of $\csc(y(x))y'(x)$ and $1$ are given by $-\tanh^{-1}\left[\cos(y(x))\right]$ and $x$, so these two must differ by a constant, say $c$.
$$-\tanh^{-1}\left[\cos(y(x))\right]=x+c$$
We can now solve equation for $y(x)$. As always, $c$ was redefined to absorb the minus sign.
\begin{align*}
\tanh^{-1}\left[\cos(y(x))\right] &= c-x\\
\cos(y(x)) &= \tanh(c-x)\\
y(x) &= 2\pi k\pm\cos^{-1}\left[\tanh(c-x)\right]
\end{align*}
In the last line, we made use of the not-so-easy-to-prove fact that for $|y|\leq 1$,
$$y=\cos(x)\iff x=2\pi k\pm\cos^{-1}(y)$$
In summary, the solutions are
$$y(x)=\pi k$$
and
$$y(x) = 2\pi k\pm\cos^{-1}\left[\tanh(c-x)\right]$$
