Convergence of $\int^{\pi/2}_0 x\sqrt{\sec x}dx$

Question

At $x=\pi/2$, $\sec x$ goes to infinity, and $x$ is fixed, so $x\sqrt{\sec x}$ goes to infinity. It seems to diverge, but the solution says it converges. I don't know how to prove it. I cannot find the antiderivative of this function or suitable functions to apply comparison test. Any help, thanks!

Answer 1

Hint: $\frac x {\sqrt {cos x}}=\frac x {\sqrt {\sin (\frac {\pi} 2 -x)}}$. Make the change of variable $y=\frac {\pi} 2 -x$ and use the fact that $\sin y \sim y$ for $y$ near $0$.

Answer 2

Composing Taylor series close to the bounds, we have $$x\sqrt{\sec (x)}=x+\frac{x^3}{4}+O\left(x^5\right)$$ $$x\sqrt{\sec (x)}=\frac{\pi }{2 \sqrt{\frac{\pi }{2}-x}}-\sqrt{\frac{\pi }{2}-x}+O\left(\left(x-\frac{\pi }{2}\right)^{3/2}\right)$$ In fact, going to higher order for the second series gives a good approximation of the definite integral (which does not show a closed form).