Definite Integration
Special Integral Techniques
Grade 12
Question:
<p>The value of the definite integral <span class="math">\int_0^{\pi/3} \ln(1 + \sqrt{3}\tan x) dx</span> equals</p>
<p>(a) <span class="math">\frac{\pi}{3}\ln 2</span></p>
<p>(b) <span class="math">\frac{\pi}{3}</span></p>
<p>(c) <span class="math">\frac{\pi}{6}\ln 2</span></p>
<p>(d) <span class="math">\frac{\pi}{2}\ln 2</span></p>
Step-by-Step Solution
Key Concept: Use the property that for integrals of the form ∫₀ᵃ f(x)dx, we can often leverage substitution and symmetry. Here, substitute tan(x) using a complementary angle approach or use the King's property by evaluating I and I' where we replace x with (π/3 - x).
<p><strong>Step 1:</strong> Let I = ∫₀^(π/3) ln(1 + √3 tan x) dx</p><p><strong>Step 2:</strong> Apply King's property by substituting x → (π/3 - x), so dx → -dx. When x = 0, the upper limit becomes π/3, and when x = π/3, the lower limit becomes 0.</p><p><strong>Step 3:</strong> We get I = ∫₀^(π/3) ln(1 + √3 tan(π/3 - x)) dx</p><p><strong>Step 4:</strong> Using the tangent subtraction formula: tan(π/3 - x) = (tan(π/3) - tan x)/(1 + tan(π/3)tan x) = (√3 - tan x)/(1 + √3 tan x)</p><p><strong>Step 5:</strong> Therefore: 1 + √3 tan(π/3 - x) = 1 + √3 · (√3 - tan x)/(1 + √3 tan x) = (1 + √3 tan x + 3 - √3 tan x)/(1 + √3 tan x) = 4/(1 + √3 tan x)</p><p><strong>Step 6:</strong> So I = ∫₀^(π/3) ln[4/(1 + √3 tan x)] dx = ∫₀^(π/3) [ln 4 - ln(1 + √3 tan x)] dx</p><p><strong>Step 7:</strong> This gives: I = ln 4 · (π/3) - I, where the second integral is our original I.</p><p><strong>Step 8:</strong> Therefore: 2I = (π/3)ln 4 = (π/3)ln 2²</p><p><strong>Step 9:</strong> Solving: 2I = (2π/3)ln 2, so I = (π/3)ln 2</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a