Compute $\int\frac{1}{3+\cos^3{x}}\mathrm{d} x$ I have an integral which seems hard for me:
$$\int\frac{1}{3+\cos^3{x}}\,\mathrm{d}x.$$
If I use Weierstrass substitution I get $I=\int{\frac{(1+t^2)^2}{3(1+t^2)^3+(1-t^2)^3}dt} $ I was stuck here
 A: I outline some steps for you to work with.
First solve
$$
t^3+3=0.
$$
Call the solutions $t_1$, $t_2$ and $t_3$ (two of them will be complex).
Next, you use the fact (partial fraction decomposition) that
$$
\begin{aligned}
\frac{1}{(x-t_1)(x-t_2)(x-t_3)}&=\frac{1}{(t_1-t_2)(t_1-t_3)(x-t_1)}+\frac{1}{(t_2-t_1)(t_2-t_3)(x-t_2)}\\
&\qquad+\frac{1}{(t_3-t_1)(t_3-t_2)(x-t_3)}.
\end{aligned}
$$
You are now about to tackle three integrals of the form (forgetting constants)
$$
\int\frac{1}{\cos x-t}\,dx.
$$
Via $u=\tan(x/2)$ you will find that
$$
\int\frac{1}{\cos x-t}\,dx=\frac{2\text{arctanh}\,\Bigl(\frac{\sqrt{1+t}\tan(x/2)}{\sqrt{1-t}}\Bigr)}{\sqrt{1-t^2}}+C
$$
Of course, in the end you should be a bit careful since the constants are complex and you expect a real result, but in principle it should work out.
A: Robert, observe that$\frac{1}{3+cos^2x}=\frac{sec^2x}{3sec^2x+1}=\frac{sec^2x}{4+3tan^2x}$ Now do u-sub $tanx=u$ on the integral and the result is essentially an $arctan$ No need for Weierstrass, just High School Calculus 
