Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>Evaluate the following: <br> 25. \(\int_0^{\pi/2} \cos 2x \cdot \log(1 + \tan x) \, dx\)</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a-x)dx, then add the original and transformed integrals to simplify the logarithmic term using trigonometric identities.
<p><strong>Step 1:</strong> Let I = ∫₀^(π/2) cos(2x)·log(1+tan x) dx</p><p><strong>Step 2:</strong> Apply the property I = ∫₀^(π/2) cos(2(π/2-x))·log(1+tan(π/2-x)) dx</p><p><strong>Step 3:</strong> Simplify using tan(π/2-x) = cot x and cos(π-2x) = -cos(2x):<br/>I = ∫₀^(π/2) -cos(2x)·log(1+cot x) dx</p><p><strong>Step 4:</strong> Rewrite log(1+cot x) = log((1+tan x)/tan x) = log(1+tan x) - log(tan x)</p><p><strong>Step 5:</strong> Add original equation to the transformed equation:<br/>2I = ∫₀^(π/2) cos(2x)·[log(1+tan x) - log(1+cot x)] dx<br/>2I = ∫₀^(π/2) cos(2x)·log(tan x) dx</p><p><strong>Step 6:</strong> Split: 2I = ∫₀^(π/2) cos(2x)·log(sin x) dx - ∫₀^(π/2) cos(2x)·log(cos x) dx</p><p><strong>Step 7:</strong> Using symmetry and known results for these integrals:<br/>∫₀^(π/2) cos(2x)·log(sin x) dx = ∫₀^(π/2) cos(2x)·log(cos x) dx (by substitution x→π/2-x)</p><p><strong>Step 8:</strong> Therefore 2I = 0</p><p>∴ <strong>Answer: 0</strong></p>
Correct Answer: 0