Show that $\int ^2_1\int^x_{\sqrt{x}}\sin \frac{\pi x}{2y}dydx+\int^4_2\int^2_{\sqrt{x}}\sin \frac{\pi x}{2y}dydx=\frac{4(\pi+2)}{\pi^3}$ Show that $$\int ^2_1\int^x_{\sqrt{x}}\sin \frac{\pi x}{2y}dydx+\int^4_2\int^2_{\sqrt{x}}\sin \frac{\pi x}{2y}dydx=\frac{4(\pi+2)}{\pi^3}$$
I sketched out the domain of the integration, it seems these two part can not be combined together. 
I tried change the order of the integration. However, even for the second part, which is easier when change order, is not easy to calculate. Like
$$\int^4_2\int^2_{\sqrt{x}}\sin \frac{\pi x}{2y}dydx=\int^2_{\sqrt{2}}[\int^{y^2}_2\sin\frac{\pi x}{2y}dx]dy=\int^2_{\sqrt{2}}\frac{-2y}{\pi}(\cos \frac{\pi y}{2}-\cos\frac{\pi}{y})dy$$
still not easy to compute... Any other methods? Thanks~
 A: Well, changing the order of integration in both cases and calculating the integral in x you're going to find something like that:
$-\int_1^2 \frac{2y}{\pi}$ cos$(\frac{\pi}{y}) \,dy$ + $\int_1^2 \frac{2y}{\pi}$ cos$(\frac{\pi}{2}) \,dy + \int_1^{\sqrt{2}} \frac{2y}{\pi}$ cos$(\frac{\pi}{y}) \,dy - \int_1^{\sqrt{2}} \frac{2y}{\pi}$ cos$(\frac{\pi y}{2}) \,dy - \int_{\sqrt{2}}^2 \frac{2y}{\pi}$ cos$(\frac{\pi y}{2}) \,dy + \int_{\sqrt{2}}^2 \frac{2y}{\pi}$ cos$(\frac{\pi}{y}) \,dy$
Note that the second integral is zero and you can sum up integral number 1 with the third and the last integral and that is equal to zero.
Now you make a change of variables and call $u = \frac{\pi y}{2}$, then $\frac{2}{\pi}\,du = \,dy$, and then you have:
$-\frac{8}{\pi^3}\int_1^{2}u$ cos $u \,du$, then you integrate by parts and you get your result.
A: You can actually combine the graphs. If you sketch out the graph you can easily see that you can integrate the area w.r.t to the y axis, as the 2 seperate areas are combined.
You can get
$$\int^2_1 \int^{y^2}_ysin\frac{\pi x}{2y}dx dy$$
Hopefully from the limits inserted it's easy to see the integrating region. After this it's a simple double integral with integration by parts as the second integral.
