Problem 34 - Essential Calculus

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{\pi / 2} \sec \theta d \theta $$

Step-by-Step Solution

Answer

The improper integral \(\int_{0}^{\pi / 2} \sec \theta d \theta\) diverges

Step 1: Set up the limit

Because the integral is improper and \(\sec(\theta)\) approaches infinity as \(\theta\) approaches \(\pi / 2\), we can rewrite the integral as: \(\lim_{b \to \frac{\pi}{2}^-} \int_0^b \sec(\theta) d\theta\)

Step 2: Evaluate the integral

The integral of the secant function is the natural logarithm of the absolute value of the secant plus the tangent; you may need to use trigonometric identities to prove this. Applying this gives us: \(\lim_{b \to \frac{\pi}{2}^-} \left[ \ln |\sec(\theta) + \tan(\theta)| \right]_0^b\)

Step 3: Substitute the limits

Substitute \(b\) and \(0\) into the expression to get: \(\lim_{b \to \frac{\pi}{2}^-} (\ln |\sec(b) + \tan(b)| - \ln |\sec(0) + \tan(0)|)\).

Step 4: Simplify the expression

Simplify the expression to get: \(\lim_{b \to \frac{\pi}{2}^-} \ln |\sec(b) + \tan(b)| \). This is because sec(0) is 1 and tan(0) is 0, yielding ln(1), which is 0. When \(b\) approaches \(\frac{\pi}{2}^- \), \sec(b) approaches positive infinity, and \tan(b) approaches positive infinity. The sum inside the absolute value is therefore positive infinite.

Step 5: Evaluate the limit

As both \sec(b) and \tan(b) approach positive infinity as \(b\) approaches \(\pi / 2\), ln|\sec(b) + \tan(b)| approaches infinity. Therefore, the improper integral diverges