<p>\(I_n = \int_0^{\pi/4} \tan^n x\, dx\), then \(\lim_{n \to \infty} n(I_n + I_{n+2})\) equals</p>
Step-by-Step Solution
Key Concept: Use the reduction formula for ∫tan^n x dx: I_n + I_(n+2) = ∫tan^n x(1 + tan²x)dx = ∫tan^n x·sec²x dx = [tan^(n+1)x/(n+1)] from 0 to π/4, which gives I_n + I_(n+2) = 1/(n+1). Then analyze the limit as n→∞.
<p><strong>Step 1: Establish the reduction formula</strong></p><p>I_n + I_(n+2) = ∫₀^(π/4) tan^n x dx + ∫₀^(π/4) tan^(n+2) x dx</p><p>= ∫₀^(π/4) tan^n x(1 + tan²x) dx</p><p>= ∫₀^(π/4) tan^n x · sec²x dx</p><p><strong>Step 2: Integrate using substitution</strong></p><p>Let u = tan x, then du = sec²x dx</p><p>I_n + I_(n+2) = ∫₀¹ u^n du = [u^(n+1)/(n+1)]₀¹ = 1/(n+1)</p><p><strong>Step 3: Find the limit</strong></p><p>lim(n→∞) n(I_n + I_(n+2)) = lim(n→∞) n · 1/(n+1)</p><p>= lim(n→∞) n/(n+1)</p><p>= lim(n→∞) 1/(1 + 1/n)</p><p>= 1</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: A