Proof $\int\limits_{0}^{\pi/2} (\cos(x))^{-2/3}\ dx$ converges I have been trying to show $\int\limits_0^{\pi/2} (\cos(x))^{-2/3}\ dx$ converges. I've been trying to bound it below by some divergent term, or bound it above by something convergent, but so far the only thing I've come up with that is that $\int\limits_0^{\pi/2} \frac{\sin(x)}{\cos(x)^{2/3}}\ dx$ converges, but I cannot figure out how it can help me in this case.
 A: Since
$x \ge \sin(x)
\ge \dfrac{2x}{\pi}
$
for $0 \le x \le \dfrac{\pi}{2}$,
$\begin{array}\\
\int\limits_0^{\pi/2} (\cos(x))^{-2/3}\ dx
&=\int\limits_0^{\pi/2} (\sin(x))^{-2/3}\ dx\\
&=\int\limits_0^{\pi/2} \dfrac{dx}{\sin^{2/3}(x)}\\
&\le\int\limits_0^{\pi/2} \dfrac{dx}{(\dfrac{2x}{\pi})^{2/3}}\\
&=(\dfrac{\pi}{2})^{2/3}\int\limits_0^{\pi/2} \dfrac{dx}{x^{2/3}}\\
&=(\dfrac{\pi}{2})^{2/3}\dfrac{x^{1/3}}{1/3}\big|_0^{\pi/2}\\
&=3(\dfrac{\pi}{2})^{1/3}\\
\end{array}
$
This works for
any power less that 1.
A: An alternative is
$$ \int_{0}^{\pi/2}\cos(x)^{-2/3}\,dx = \int_{0}^{\pi/2}\sin(x)^{-2/3}\,dx = \int_{0}^{1}\frac{dz}{z^{2/3}\sqrt{1-z^2}}=\frac{1}{2}\int_{0}^{1}u^{-5/6}(1-u)^{-1/2}\,du$$
and by the properties of the Beta function and the relations between $\Gamma$ and the $\text{AGM}$ the RHS is
$$ \frac{\Gamma(1/6)\,\Gamma(1/2)}{2\,\Gamma(2/3)}=\sqrt{\frac{3}{16\pi}}\,\Gamma(1/6)\,\Gamma(1/3)= \frac{2^\frac{1}{3}\cdot 3^{\frac{3}{4}}\cdot \pi}{ \operatorname{AGM}\left(2,\sqrt{2+\sqrt{3}}\right)^{\frac13}\operatorname{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^\frac23}. $$
A: Basically, you only have to care about the"bad" points, and in this case, $x=\pi/2$. Now, expand $(\cos x)^{-2/3}$ around $\pi/2$, you get 
$$
(\cos x)^{-2/3} = (\sin (\pi/2-x))^{-2/3}=((\pi/2-x)+O((\pi/2-x)^3))^{-2/3}=(\pi/2-x)^{-2/3}(1+O((\pi/2-x)^2)
$$
So it's bounded by say $2(\pi/2-x)^{-2/3}$ in a tiny interval $[\pi/2-\epsilon, \pi/2)$, and you know $2(\pi/2-x)^{-2/3}$ is integrable over $[\pi/2-\epsilon, \pi/2)$. 
