Why is the following integral improper? $\int_0^\frac {\pi}{2} sec(x) dx$ I'm asked to explain why the following integral is improper and determine whether the integral is convergent or divergent. I really am not sure how to do these problems and I am unsure on where to start with this. The integral is:
$\int_0^\frac {\pi}{2} sec(x) dx$
Can someone explain to me how I approach these sort of problems?
Thanks
 A: A definite integral is improper if either the interval you're integrating over is infinite (for example: $\int_1^\infty \frac{1}{x^2} dx$) or if at least one point in the interval you're integrating over is not in the domain of the integrand. In this case, the latter is true; there's a number $c$ in $\lbrack 0, \frac{\pi}{2}\rbrack$ at which $\sec x = 1/\cos x$ is undefined. Do you see where it is? Conveniently there's only one such $c$ in this exercise.
Once you know where $c$ is, the improper integral is defined to be
$$
    \lim_{t \to c^-}\int_0^c \sec(x) dx + \lim_{t\to c^+}\int_c^{\pi/2} \sec x dx.
$$
if $0 < c < \pi/2$,
$$
    \lim_{t\to 0^+}\int_c^{\pi/2} \sec x dx
$$
if $c = 0$, and
$$
    \lim_{t \to (\pi/2)^-}\int_0^c \sec(x) dx
$$
if $c = \pi/2$, where there's a different definition if $c$ equals one of the end points, since otherwise one of the integrals in the first sum doesn't make sense.
A: From Wolfram MathWorld:

An improper integral is a definite integral that has either or both
  limits infinite or an integrand that approaches infinity at one or
  more points in the range of integration.

Are either of the limits infinite? Have you tried plotting this function? Are there any vertical asymptotes? What methods would you use to determine convergence or divergence?
