I entered this integration problem to Mathematica Online Integrator an got a solution I would never have been able to find manually.
$$\int\root 3 \of{\cos(x)^2}\,dx=\frac{(-3\cos(x)\root 3 \of{\cos(x)^2}*Hypergeometric2F1[1/2, 5/6, 11/6, \cos(x)^2]*\sin(x))}{5\sqrt{\sin(x)^2}}$$
I would like to see/learn how the trick works to solve/reproduce this the manual way.
Expanding the above comment, we can write (say, for $x\in\left[0,\frac{\pi}{2}\right]$) \begin{align} \int \cos^{r}x\,dx&=-\int \cos^r x\left(1-\cos^2 x\right)^{-\frac12}\,d(\cos x)=\\ &=-\int \sum_{k=0}^{\infty}\frac{\Gamma(k+\frac12)}{k!\,\Gamma(\frac12)}\cos^{r+2k}x\,d(\cos x)=\\ &=-\sum_{k=0}^{\infty}\frac{\Gamma(k+\frac12)}{k!\,\Gamma(\frac12)}\frac{\cos^{r+1+2k}x}{r+1+2k}=\\ &=-\frac{\cos^{r+1}x}{r+1}\sum_{k=0}^{\infty}\frac{(\frac12)_k(\frac{r+1}{2})_k}{k!\,(\frac{r+3}{2})_k}\left(\cos^2 x\right)^k=\\ &=-\frac{\cos^{r+1}x}{r+1}{}_2F_1\left(\frac12,\frac{r+1}{2};\frac{r+3}{2};\cos^2 x\right),\tag{1} \end{align} where $(a)_k=\frac{\Gamma(a+k)}{\Gamma(a)}$ denotes the Pochhammer symbol. It remains to set $r=\frac23$ in the last formula.