Compute the flux of the vector field

$\displaystyle xy\sin(z)\vec{i}+\cos(xz)\vec{j}+y\cos(z)\vec{k}$

across the surface of the ellipsoid $\displaystyle \frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}=1$

Attempt:
I was thinking of making the following substitutions:

$\displaystyle x=\frac{\sin(\phi)\cos(\theta)}{2}$

$\displaystyle y=\frac{\sin(\phi)\sin(\theta)}{3}$

$\displaystyle z=\frac{cos(\phi)}{4}$

now calculating $\displaystyle (r_\phi \times r_\theta)$ gives:

$\displaystyle \frac{\sin^2(\phi)\cos(\theta)}{12}\vec{i}+\frac{\ sin^2(\phi)\sin(\theta)}{8}\vec{j}+\frac{\sin(\phi )\cos(\phi)}{6}\vec{k}$

now putting everything into $\displaystyle \iint_S F\cdot(r_\phi \times r_\theta) \ dS$

gives me:

$\displaystyle \int_{\theta=0}^{\theta=2\pi}\int_{\phi=0}^{\phi=\ pi} \left(\frac{\sin(\phi)\cos(\theta)}{2}\right)\left (\frac{\sin(\phi)\sin(\theta)}{3}\right)\sin\left( \frac{cos(\phi)}{4}\right)\cdot\left(\frac{\sin^2( \phi)\cos(\theta)}{12}\right)$$\displaystyle {\color{blue}+\cos\left(\left(\frac{\sin(\phi)\cos (\theta)}{2}\right)\left(\frac{cos(\phi)}{4}\right )\right)\cdot\left(\frac{\sin^2(\phi)\sin(\theta)} {8}\right)}$$\displaystyle +\left(\frac{\sin(\phi)\sin(\theta)}{3}\right)\cos \left(\frac{cos(\phi)}{4}\right)\cdot\left(\frac{\ sin(\phi)\cos(\phi)}{6}\right) \ d\phi \ d\theta$

now everything except for the part in blue is integrable, where the first and third part give 0, is there a simpler way of approaching this problem?