Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?

Question

Is $\int_{-\pi/2}^{\pi/2}\sec(x)$ bounded?

It seems like it shouldn't be, since:

$$\int \sec(x) = \ln |\sec(x) + \tan(x)|+C$$

and $\sec(x)\to\infty$ as $x \to (\pi/2)^-$. But, I know integrals can sometimes be counter-intuitive and it may be possible for an integral to be bounded, even if the function itself goes to infinity.

Can anyone confirm?

Answer 1

Correct, the integral diverges to $\infty.$

Answer 2

The integrand is even, so your integral is $2\lim_{x\to(\pi/2)^-}[\ln|\sec x+\tan x|]_0^x=\infty$.

Answer 3

Note that$$\lim_{x\to\pi/2}\frac{\sec(x)}{\frac1{\pi/2-x}}=1.$$So, since the integral $\int_0^{\pi/2}\frac1{\pi/2-x}\,\mathrm dx$ diverges, then so does $\int_0^{\pi/2}\sec(x)\,\mathrm dx$.