This is what I have, though I'm most likely missing something:
$$\int\limits^\pi_1\frac{\sin\left(\frac{1}{\cos x}\right)}{\sqrt{x} }dx \le \int\limits_1^{\pi}\sin\left(\frac{1}{\cos x}\right)dx\le\int\limits_1^\pi dx=\pi-1$$
Therefore converges.
What about when $x=\frac{\pi}{2}\implies \cos x=0$, what to do about this?
Best Answer
On the interval $[1,\pi]$:
$$\left|\frac{\sin\left(\frac{1}{\cos x}\right)}{\sqrt{x}}\right|\le\left|\sin\left(\frac{1}{\cos x}\right)\right|\le 1$$
therefore
$$ \int\limits^\pi_1 \left| \frac{\sin\left(\frac{1}{\cos x}\right)}{\sqrt{x} }\right|\,dx\le \int\limits^\pi_1 \,dx=\pi - 1$$
so the integral converges.