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.

Draw a diagram showing the bounded region & change order of integration as follows $$\int_{0}^{\frac{\pi^3}{8}}\ dx\int_{\sqrt[3]{x}}^{\frac{\pi}{2}}\cos\left(\frac{2x}{\pi y}\right)\ dy$$
$$=\int_{0}^{\frac{\pi}{2}}\ dy\int_{0}^{y^3}\cos\left(\frac{2x}{\pi y}\right)\ dx$$
$$=\int_{0}^{\frac{\pi}{2}}\ dy\left(\frac{\pi y}{2}\sin\left(\frac{2x}{\pi y}\right)\right)_{0}^{y^3}$$
$$=\frac{\pi }{2}\int_{0}^{\frac{\pi}{2}}y\sin\left(\frac{2y^2}{\pi }\right)\ dy$$
$$=\frac{\pi }{2}\frac{\pi }{4}\int_{0}^{\frac{\pi}{2}}\sin\left(\frac{2y^2}{\pi }\right)\ d\left(\frac{2y^2}{\pi}\right)$$
$$=\frac{\pi^2 }{8}\left(-\cos\left(\frac{2y^2}{\pi }\right)\right)_{0}^{\frac{\pi}{2}}$$
$$=\frac{\pi^2 }{8}\left(-\cos\left(\frac{\pi}{2 }\right)+\cos 0\right)=\color{red}{\frac{\pi^2}{8}}$$
