What is integration of $\csc (\pi \sqrt y )$ I have given an O.D.E 

$$\frac{dy}{dt}=\sin(\pi\sqrt y)$$

We can solve this by doing $$\frac{1}{\sin(\pi\sqrt y) }\;dy=dt$$ and then integrating each side 
$$\int\frac{1}{\sin(\pi\sqrt y) }\;dy=\int dt$$
$$\int \csc(\pi\sqrt y)\;dy= \int dt$$
after that i put $\pi\sqrt y=z$, $\;$I am getting eqaution like $$\frac{2}{\pi^{2}}\int {\csc(z).z}=\int dt$$
further this i am not able to solve I applied the method of integration by parts but not getting anywhere 
Please help
Thnkyou
 A: I re-edited the answer, with a slightly different substitution, to better reconstruct W.A. solution.
First re-write your integral as
\begin{eqnarray}
\mathcal I = \int \frac{x}{\sin x} dx = \int \frac{2ix}{e^{ix}\left(1-e^{-2ix}\right)}dx.
\end{eqnarray}
Then use the substitution $e^{ix} = u$, which yields
$$\mathcal I = 2i \int \frac{\log u}{1-u^2}du = i\int\frac{\log u}{1-u}du+i\int\frac{\log u}{1+u}du.$$
Recalling the definition of the dilogarithm for the first term, and integration by parts for the second term gives you
$$\mathcal I = i\operatorname{Li}_2(1-u) + i\log u \log(1+u) - i\int\frac{\log(1+u)}{u}du.$$
Using again the dilogarithm to integrate the last term brings us to
$$\mathcal I = i\operatorname{Li}_2(1-u) + i\log u \log(1+u)+\operatorname{Li}_2(-u)+C,$$
and substituting back,
$$\mathcal I = i\operatorname{Li}_2\left(1-e^{ix}\right) -x\log\left(1+e^{ix}\right) +i\operatorname{Li}_2\left(-e^{ix}\right) + C.\tag{1}\label{eq1}$$
Using the the reflection formula
$$\operatorname{Li}_2(1-z)+ \operatorname{Li}_2(z) = \frac{\pi^2}6 -\log z \log (1-z)$$
to replace the first term in \eqref{eq1}, you obtain the W.A. expression, as given in J.G.'s comment, that is
$$\mathcal I = x\log\left(1-e^{ix}\right)-x\log\left(1+e^{ix}\right)+i\left[\operatorname{Li}_2\left(-e^{ix}\right)-\operatorname{Li}_2\left(e^{ix}\right)\right]+C_1.$$
