Find $ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta$ Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta  $$
Question- Can I take the limit inside the integral?
My try-
$$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left(\frac{1+2r^2}{1-r^2\cos2\theta} \right)^{1/3}\, d\theta  $$
$$ I= \int_{-\pi}^{\pi}   \lim_{r \to 1^{-}}  (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta  $$
$$ I=3^{1/3} \int_{-\pi}^{\pi}  \frac{1}{(1-\cos2\theta)^{1/3}}d\theta  $$
$$ I= 3^{1/3}2\int_{0}^{\pi}    \frac{1}{(1-\cos2\theta)^{1/3}}   d\theta       $$
$$ I= 3^{1/3}4\int_{0}^{\pi/2}    \frac{1}{(1-\cos2\theta)^{1/3}}   d\theta       $$
$$ I= (3/2)^{1/3}4\int_0^{\pi/2} 
\sin^{-2/3}\theta  \  d\theta       $$
$$I= 4(3/2)^{1/3} \frac{\Gamma(1/6)\Gamma(1/2) }{\Gamma(2/3)}. $$

This is not an unswer, just a hint and too long for a comment: The integral above may be transformed with  $\theta = \frac{\arccos(y)}{2}$
$$2\, \left(1+2\, r^2\right)^{1/3} \int_{-1}^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dy$$
Next we can split the integral in two parts with $p=-y$ in the first one:
$$2\,\left(1+2\, r^2\right)^{1/3} \left(\int_0^1 \left(1+r^2 p\right)^{-1/3} \left(1-p^2\right)^{-\frac{1}{2}} \, dp +  \int_0^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dp \right) $$
and evaluate them in form of hypergeometric functions. For this step e.g. we can express the integrand in form of MeijerG - Function. The first integral will be transformed with $q = p^2$  apply the formular Mathematical Functions Site and simplify:
$$I_1 = \frac{\left(1+2\, r^2\right)^{1/3}}{\gamma\left[\frac{1}{3}\right]} \int_0^1 (1-q)^{-\frac{1}{2}} q^{-\frac{1}{2}} \,G_{1,1}^{1,1}\left( \sqrt{q}\, r^2\left\vert 
\begin{array}{c}
\frac{2}{3}\\ 
0%
\end{array}%
\right. \right)\,
\, dq$$
$$ = \left(1+2 \,r^2\right)^{1/3} \left(\pi  \,_{2}F_{1}\left[\frac{1}{6},\frac{2}{3},1,r^4\right]-\frac{2}{3} r^2 \,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\
\left\{\frac{3}{2},\frac{3}{2}\right\},r^4\right]\right)$$
The second integral $I_2$ is performed in the same way. The limit $r \to 1$ Mathematical Functions Site and Mathematical Functions Site exists e.g. for $I_1$
$$\lim_{r \to 1}\,I_1=\frac{\frac{3 \pi\,  \gamma\
\left[\frac{1}{6}\right]}{\gamma\left[\frac{1}{3}\right]\, \gamma\left[\frac{5}{6}\right]}-2\,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\left\{\frac{3}{2},\frac{3}{2}\right\},1\right]}{3^{2/3}}$$
and for  $I_2$ in the same way. The sum of $I = I_1  + I_2$
delivers just the half of your answer:
$$I= \frac{2^{1/3}\, 3^{5/6}\, \gamma\left[\frac{1}{3}\right]^2}{\gamma\left[\frac{2}{3}\right]} $$.
Therefore, inspite of a calculation error of factor 2, @user1055 is right, the limit may be also calculated inside the integral.
