Definite Integration
King's property of definite integrals
Grade 12
Question:
<p>Evaluate \(I = \int_{\pi/6}^{\pi/3} \dfrac{dx}{1 + \sqrt{\tan x}}\).</p>
<p>\(\dfrac{\pi}{6}\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{12}\)</p>
Step-by-Step Solution
Key Concept: Use the substitution property: if I = ∫[a,b] f(x)dx, then I = ∫[a,b] f(a+b-x)dx. Adding both forms eliminates the problematic √tan x term through a trigonometric identity.
<p><strong>Step 1:</strong> Let I = ∫[π/6 to π/3] dx/(1 + √tan x)</p><p><strong>Step 2:</strong> Use property: I = ∫[π/6 to π/3] f(π/6 + π/3 - x)dx = ∫[π/6 to π/3] dx/(1 + √tan(π/2 - x))</p><p><strong>Step 3:</strong> Since tan(π/2 - x) = cot x, we have √tan(π/2 - x) = √(cot x) = 1/√(tan x)</p><p><strong>Step 4:</strong> So I = ∫[π/6 to π/3] dx/(1 + 1/√tan x) = ∫[π/6 to π/3] √tan x/(√tan x + 1) dx</p><p><strong>Step 5:</strong> Adding the two expressions for I:</p><p>2I = ∫[π/6 to π/3] [1/(1 + √tan x) + √tan x/(1 + √tan x)] dx = ∫[π/6 to π/3] dx = π/3 - π/6 = π/6</p><p><strong>Step 6:</strong> Therefore I = π/12</p><p>∴ Answer: D</p>
Correct Answer: D