Prove: $\int_0^{\pi/2} \cos(\sec^{\;p} x) dx$ converges, $p > 0$.

Question

Prove that $\int_0^{\pi/2} \cos(\sec^{\;p} x) dx$ converges for all $p>0$.

I tried to do a change of variables through $u = \sec^{\;p} x$, but it is not sufficient to use the comparison test later.

Is there a better approach?

Answer 1

HINT:

Let $y=\sec^p (x)$ so that $x=\arctan\left(\sqrt{y^{2/p}-1}\right)$. Then, $dx=\frac{1}{py\,\sqrt{y^{2/p}-1}}\,dy$. Therefore, we have for $p>0$

$$\int_0^{\pi/2} \cos (\sec^p(x))\,dx=\int_1^\infty \frac{\cos (y)}{py\sqrt{y^{2/p}-1}}\,dy$$

Can you finish now?

Answer 2

Since $|cos(sec^{\;p} x)| \leq 1$, and $\int_0^{\pi/2} 1 dx$ converges, by the comparison test you can verify directly that your integral has absolute convergence.