82-88. Improper integrals Evaluate the following integrals or show that the integral diverges.
84. ∫ (from 0 to π) sec²x dx*(Note: Potential improperness at x = π/2)*

Verified step by step guidance

1. Identify the integral to evaluate: \(\int_0^{\pi} \sec^2 x \, dx\). Notice that \(\sec^2 x = \frac{1}{\cos^2 x}\), and the function has a potential vertical asymptote where \(\cos x = 0\), which occurs at \(x = \frac{\pi}{2}\) within the interval \([0, \pi]\).
2. Since the integrand is not defined at \(x = \frac{\pi}{2}\), split the integral into two improper integrals at this point: \(\int_0^{\pi} \sec^2 x \, dx = \int_0^{\frac{\pi}{2}} \sec^2 x \, dx + \int_{\frac{\pi}{2}}^{\pi} \sec^2 x \, dx\).
3. Rewrite each integral as a limit to handle the improper behavior: \(\int_0^{\frac{\pi}{2}} \sec^2 x \, dx = \lim_{t \to \frac{\pi}{2}^-} \int_0^t \sec^2 x \, dx\) and \(\int_{\frac{\pi}{2}}^{\pi} \sec^2 x \, dx = \lim_{s \to \frac{\pi}{2}^+} \int_s^{\pi} \sec^2 x \, dx\).
4. Find the antiderivative of \(\sec^2 x\), which is \(\tan x\), and express each integral in terms of \(\tan x\): \(\int \sec^2 x \, dx = \tan x + C\).
5. Evaluate the limits for each integral using the antiderivative: compute \(\lim_{t \to \frac{\pi}{2}^-} (\tan t - \tan 0)\) and \(\lim_{s \to \frac{\pi}{2}^+} (\tan \pi - \tan s)\), then analyze whether these limits converge or diverge to determine the behavior of the original integral.

Key Concepts

Improper Integrals

Improper integrals occur when the interval of integration is infinite or the integrand has an infinite discontinuity within the interval. To evaluate them, one must use limits to approach the problematic point and determine if the integral converges or diverges.

Behavior of sec²x and its Discontinuities

The function sec²x = 1/cos²x has vertical asymptotes where cos x = 0, such as at x = π/2. Understanding these discontinuities is crucial because they can cause the integral to be improper and affect convergence.

Evaluating Limits in Definite Integrals

When an integral is improper due to a discontinuity inside the interval, it is split at the point of discontinuity. Each part is evaluated as a limit approaching the discontinuity to check if the integral converges or diverges.