<p>The value of \(\int_0^{\pi} \sec^2 x\, dx\) is</p>
Step-by-Step Solution
Key Concept: The antiderivative of sec²x is tan(x), and you must carefully evaluate tan(x) at the boundary points π and 0, recognizing that tan(π) = 0 and tan(0) = 0, making this integral problematic at the boundaries.
<p><strong>Step 1:</strong> Identify the antiderivative: ∫sec²x dx = tan(x) + C</p><p><strong>Step 2:</strong> Check for discontinuities in [0, π]: sec(x) = 1/cos(x) is undefined at x = π/2 where cos(π/2) = 0</p><p><strong>Step 3:</strong> Since there's a vertical asymptote at x = π/2 within the integration interval, this becomes an improper integral: ∫₀^(π/2⁻) sec²x dx + ∫_(π/2⁺)^π sec²x dx</p><p><strong>Step 4:</strong> The first integral: lim_{t→π/2⁻} [tan(x)]₀^t = lim_{t→π/2⁻} tan(t) = +∞</p><p><strong>Step 5:</strong> Since the integral diverges to infinity, the definite integral <strong>does not exist (diverges)</strong>.</p><p>∴ Answer: D (The integral diverges/does not exist)</p>
Correct Answer: D