Double integral and polar coordinates Please, help me solve this double integral
$$\int^{2\pi}_0d\varphi\int^{2}_1\frac{1}{\sqrt{\rho^3\cos^3\varphi+\rho^3\sin^3\varphi}}\rho\,d\rho$$
I really don't know how to figure out and carry of $\cos^3\varphi$ and $\sin^3\varphi$ into $d\varphi$
 A: Since $\sqrt{\rho^3\cos^3{\phi}+\rho^3\sin^3{\phi}}=\sqrt{\rho^3\left(\cos^3{\phi}+\sin^3{\phi}\right)}=\rho\sqrt{\rho}\sqrt{\cos^3{\phi}+\sin^3{\phi}}$,
$$\int_{0}^{2\pi}d\phi\int_{1}^{2}\rho\,d\rho\frac{1}{\sqrt{\rho^3\cos^3{\phi}+\rho^3\sin^3{\phi}}}\\
= \int_{0}^{2\pi}d\phi\int_{1}^{2}d\rho\frac{1}{\sqrt{\rho}\sqrt{\cos^3{\phi}+\sin^3{\phi}}}\\
=\left(\int_{1}^{2}\frac{d\rho}{\sqrt{\rho}}\right)\left(\int_{0}^{2\pi}\frac{d\phi}{\sqrt{\cos^3{\phi}+\sin^3{\phi}}}\right).$$
The integral over $\rho$ evaluates to $\int_{1}^{2}\frac{d\rho}{\sqrt{\rho}}=2(\sqrt{2}-1)$ of course. The integral over the angular coordinate, however, can't be done in elementary terms.

Postscript: there's something screwy with the angular integral that I haven't resolved. Intuitively, the integrand $\frac{1}{\sqrt{\cos^3{\phi}+\sin^3{\phi}}}$ is periodic in $\phi$ with period $2\pi$ and as such any antiderivative would be similarly $2\pi$-periodic in $\phi$; thus, integrating over a full period should result in the definite integral vanishing identically (the integral over half a period should be the negative of the integral over the other half). However, note that $\cos^3{\phi}+\sin^3{\phi}<0$ for $\frac{3\pi}{4}<\phi<\frac{7\pi}{4}$, resulting in a purely imaginary integrand within this interval. So instead of getting a positive number and a balancing negative number for the two integrals, we get a real number and an imaginary number. For the moment, I'm stumped.
