Calculation of double integral. Calculate the double integral:
$$\int_{0}^{\frac{\pi^3}{8}}\; dx \int _{\sqrt[3] x}^{\frac{\pi}{2}}\; \cos\frac{2x}{\pi y}\;dy\;.$$
Can someone hint how to approach this as we have to integrate with respect to y but y is in denominator. I think the right approach might be changing it into polar co-ordinates but I am not able to set the limits.
 A: HINT:
$$\int_{0}^{\pi^3/8}\int_{x^{1/3}}^{\pi/2}\cos\left(\frac{2x}{\pi y}\right)\,dy\,dx=\int_{0}^{\pi/2}\int_{0}^{y^3}\cos\left(\frac{2x}{\pi y}\right)\,dx\,dy \tag 1$$

SPOLIER ALERT: Scroll over the highlighted area to reveal the full solution

 We being with the integral on the right-hand side of $(1)$ and evaluate the inner integral as $$\int_{0}^{y^3}\cos\left(\frac{2x}{\pi y}\right)\,dx=\frac{\pi y}{2}\sin\left(\frac{2y^2}{\pi}\right) \tag 2$$We wrap up the integration by proceeding to integrate the right-hand side of $(2)$ over $y$.  We obtain $$\int_0^{\pi/2}\frac{\pi y}{2}\sin\left(\frac{2y^2}{\pi}\right)=\frac{\pi^2}{8}$$And we are done!

A: Change the order of integral. We have
$$\int_0^{\pi^3/8} \int_{\sqrt[3]{x}}^{\pi/2} f(x,y)dydx = \int_0^{\pi/2} \int_0^{y^3} f(x,y) dx dy$$
Hope you can finish it from here.
A: Draw a diagram showing the bounded region & change order of integration as follows $$\int_{0}^{\frac{\pi^3}{8}}\ dx\int_{\sqrt[3]{x}}^{\frac{\pi}{2}}\cos\left(\frac{2x}{\pi y}\right)\ dy$$
$$=\int_{0}^{\frac{\pi}{2}}\ dy\int_{0}^{y^3}\cos\left(\frac{2x}{\pi y}\right)\ dx$$
$$=\int_{0}^{\frac{\pi}{2}}\ dy\left(\frac{\pi y}{2}\sin\left(\frac{2x}{\pi y}\right)\right)_{0}^{y^3}$$
$$=\frac{\pi }{2}\int_{0}^{\frac{\pi}{2}}y\sin\left(\frac{2y^2}{\pi }\right)\ dy$$
$$=\frac{\pi }{2}\frac{\pi }{4}\int_{0}^{\frac{\pi}{2}}\sin\left(\frac{2y^2}{\pi }\right)\ d\left(\frac{2y^2}{\pi}\right)$$
$$=\frac{\pi^2 }{8}\left(-\cos\left(\frac{2y^2}{\pi }\right)\right)_{0}^{\frac{\pi}{2}}$$
$$=\frac{\pi^2 }{8}\left(-\cos\left(\frac{\pi}{2 }\right)+\cos 0\right)=\color{red}{\frac{\pi^2}{8}}$$
