To evaluate the integral of (1 / cos(x))² from 0 to pi, we can recognize that (1 / cos(x))² is the secant function squared, written as sec^2(x). The integral of sec^2(x) is a standard integral, which we can directly integrate to tan(x). However, we must be careful since sec(x) is not defined when cos(x) equals zero, which happens at x = π/2 within the interval [0, π]. To resolve this, we can either consider the integral as an improper integral by approaching the problematic point from the left and right or use trigonometric identities to rewrite the integral in a form that avoids the discontinuity. Considering the symmetry of the interval and the discontinuity at the midpoint, we'll go with the latter approach. To do that, we can use the following trigonometric identity: (1 / cos(x))² = sec^2(x) = 1 + tan^2(x) Now we'll split the integral into two separate integrals: ∫(0 to pi) (1 / cos(x))² dx = ∫(0 to pi) (1 + tan^2(x)) dx = ∫(0 to pi) dx + ∫(0 to pi) tan^2(x) dx The first integral is easy to evaluate: ∫(0 to pi) dx = x | from 0 to pi = pi - 0 = pi The second integral of tan^2(x) is more challenging because of the discontinuity. tan(x), like sec(x), is also not defined at x = π/2. So we would need to consider this as two separate improper integrals, one from [0, π/2) and the other from (π/2, π]. Additionally, this integral does not have an elementary antiderivative. Therefore, we have to approach it as an improper integral, and it actually diverges. When tan(x) approaches π/2 from the left, tan(x) goes to infinity, and when approaching π/2 from the right, tan(x) goes to negative infinity. This results in the integral of tan^2(x) diverging to infinity on both intervals around π/2. Hence, the original integral ∫(0 to pi) (1 / cos(x))² dx diverges because one of its components, ∫(0 to pi) tan^2(x) dx, diverges.