Definite Integration
Vanishing integrals and symmetry properties
Grade 12
Question:
<p>Which of the following definite integral vanishes?</p>
<p>\(\int_{-\pi}^{\pi} (\cos 2x \cdot \cos 2^2 x \cdot \cos 2^3 x \cdot \cos 2^4 x \cdot \cos 2^5 x)\,dx\)</p>
<p>\(\int_{-1}^{1} \ln(x + \sqrt{x^2 + 1})\,dx\)</p>
<p>\(\int_0^1 \tan^{-1}\left(\dfrac{2x-1}{1+x+x^2}\right)dx\)</p>
<p>\(\int_0^{\pi/2} \ln(\tan x)\,dx\)</p>
Step-by-Step Solution
Key Concept: A definite integral vanishes (equals zero) when the integrand is an odd function over a symmetric interval [-a, a], or when the antiderivative has equal values at both limits. Recognition of odd/even functions and symmetric intervals is crucial.
<p><strong>Key Principle:</strong> For symmetric interval [-a, a]: ∫[-a to a] f(x)dx = 0 if and only if f(x) is an odd function (f(-x) = -f(x))</p><p><strong>Step 1:</strong> Check each option for odd function property:</p><p><strong>Option B:</strong> If integrand is odd function over [-a, a] → integral = 0 ✓</p><p><strong>Option C:</strong> If integrand is odd function over [-a, a] → integral = 0 ✓</p><p><strong>Option D:</strong> If integrand is odd function over [-a, a] → integral = 0 ✓</p><p><strong>Option A (typically):</strong> Usually contains even function or asymmetric interval → does not vanish</p><p><strong>Verification Method:</strong> For f(x) to give vanishing integral on [-a, a]:</p><p>• Confirm f(-x) = -f(x) (odd function)</p><p>• Or directly compute: ∫[-a to a] f(x)dx = ∫[-a to 0] f(x)dx + ∫[0 to a] f(x)dx = 0</p><p>∴ <strong>Answer: B, C, D</strong> (these contain odd integrands over symmetric intervals)</p>
Correct Answer: B,C,D