How can I calculate $\int_1^2 \int_{\sqrt{x}}^x \sin\left(\frac{\pi x}{2y}\right) \,dy \, dx$? Good night i have problem solving this integral.
$$\int_1^2 \int_{\sqrt{x}}^x \sin\left(\frac{\pi x}{2y}\right) \,dy \, dx$$
I make the area of integration, but i cannot solve the integrat, i don't know how! please help me.

How i can make this??
I think in this form: 
$$ \sin\left(\frac{\pi x}{2y}\right)=\sqrt{\frac{1-\cos(\pi x)}{2y}}$$ but i don't know...
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{y = {\pi x \over 2t}}$:
\begin{align}
&\color{#f00}{%
\int_{1}^{2}\int_{\root{x}}^{x}\sin\pars{{\pi x \over 2y}}\,\dd y\,\dd x} =
-\,{\pi \over 2}\int_{1}^{2}x
\int_{\pi\root{x}/2}^{\pi/2}{\sin\pars{t} \over t^{2}}\,\dd t\,\dd x
\\[3mm] = &\
-\,{\pi \over 2}\int_{1}^{2}x\braces{%
\left.-\,{\sin\pars{t} \over t}\right\vert_{\pi\root{x}/2}^{\pi/2} + \int_{\pi\root{x}/2}^{\pi/2}{\cos\pars{t} \over t}\,\dd t}\,\dd x
\\[3mm] = &\
-\,{\pi \over 2}\int_{1}^{2}x\bracks{%
-\,{1 \over \pi/2} + {\sin\pars{\pi\root{x}/2} \over \pi\root{x}/2} + \mathrm{Ci}\pars{{\pi \over 2}} - \mathrm{Ci}\pars{{\pi\root{x} \over 2}}}
\,\dd x
\end{align}
where $\mathrm{Ci}$ is the Cosine Integral function:
$\ds{\mathrm{Ci}\pars{x} \equiv -\int_{x}^{\infty}{\cos\pars{t} \over t}
\,\dd t}$. Then,
\begin{align}
&\color{#f00}{%
\int_{1}^{2}\int_{\root{x}}^{x}\sin\pars{{\pi x \over 2y}}\,\dd y\,\dd x}
\\[3mm] = &\
{3 \over 2} - {3\pi \over 4}\,\mathrm{Ci}\pars{{\pi \over 2}} -
\int_{1}^{2}\root{x}\sin\pars{{\pi\root{x} \over 2}}\,\dd x +
{\pi \over 2}\int_{1}^{2}x\,\mathrm{Ci}\pars{{\pi\root{x} \over 2}}\,\dd x
\\[3mm] = &\
{3 \over 2} - {3\pi \over 4}\,\mathrm{Ci}\pars{{\pi \over 2}} -
{16 \over \pi^{3}}\int_{\pi/2}^{\root{2}\pi/2}t^{2}\sin\pars{t}\,\dd t +
{16 \over \pi^{3}}\int_{\pi/2}^{\root{2}\pi/2}t^{3}\,\mathrm{Ci}\pars{t}\,\dd t
\end{align}
The last integral can be integrated by parts such that we can use $\ds{\mathrm{Ci}'\pars{x} = {\cos\pars{x} \over x}}$. 
