Evaluate the integral I need some help in evaluating this integral:

$$\int {\frac{{\cos x}}{{{{\sin }^3}x + \sin x + 4}}dx} $$

I've tried using the substitution $u=\sin{x}$ but I ended up with a cubic polynomial in the denominator.

For anyone who would like to see the method for evaluating the indefinite integral:
$\int{\frac{\mathrm{cos}x}{\mathrm{sin}^3x + \mathrm{sin}x + 4}}\mathrm{dx} = \int{\frac{\mathrm{du}}{u^3 + u + 4}}$
Which is by the substitution $u = \mathrm{sin}x$
Then let the roots of the polynomial be $r_1, r_2,$ and $r_3$. The fraction can then be split up:
$\frac{1}{u^3 + u + 4} = \frac{1}{(u-r_1)(u-r_2)(u-r_3)} = \frac{1}{(u-r_1)(r_1-r_2)(r_1-r_3)} + \frac{1}{(u-r_2)(r_2-r_1)(r_2-r_3)} + \frac{1}{(u-r_3)(r_3-r_1)(r_3-r_2)} $
Which can be integrated quite easily as:
$\int{(\frac{1}{(u-r_1)(r_1-r_2)(r_1-r_3)} + \frac{1}{(u-r_2)(r_2-r_1)(r_2-r_3)} + \frac{1}{(u-r_3)(r_3-r_1)(r_3-r_2)}})\mathrm{dx}\\= \frac{\mathrm{log}_e(u-r_1)}{(r_1-r_2)(r_1-r_3)} + \frac{\mathrm{log}_e(u-r_2)}{(r_2-r_1)(r_2-r_3)} + \frac{\mathrm{log}_e(u-r_3)}{(r_3-r_1)(r_3-r_2)} + \mathrm{C}\\= \frac{\mathrm{log}_e(\mathrm{sin}x-r_1)}{(r_1-r_2)(r_1-r_3)} + \frac{\mathrm{log}_e(\mathrm{sin}x-r_2)}{(r_2-r_1)(r_2-r_3)} + \frac{\mathrm{log}_e(\mathrm{sin}x-r_3)}{(r_3-r_1)(r_3-r_2)} + \mathrm{C}$
Now the last thing to do is to find $r_1, r_2,$ and $r_3$. Now, I didn't actually know how to do this, so I just looked at this which I used to find the roots as approximately:
$-1.38, 0.69 + 1.56i, 0.69 - 1.56i$. 
If you want you can get the exact values by following the link (you can get the first root via this method, and then the next two can easily be deduced from a resulting quadratic).
