Definite Integration
Definite integrals using symmetry properties
Grade 12
Question:
<p>Evaluate the integral:
\[ I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)}{e^x + 1}\, dx \]</p>
<p>\(\dfrac{\pi}{2\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{4\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{3\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: Use the substitution property f(x) + f(-x) to decompose the integrand. The first term is even (simplifies to 2tan⁻¹x) and the second is odd, so the odd component vanishes over a symmetric interval.
<p><strong>Step 1:</strong> Recognize the interval [-1/√3, 1/√3] is symmetric about origin. Use the property that for symmetric intervals: I = ∫[f(x) + g(x)]/(e^x + 1) dx.</p><p><strong>Step 2:</strong> Split into two parts:<br>- Let A(x) = cos⁻¹(2x/(1+x²)) and B(x) = tan⁻¹(2x/(1-x²))<br>- Note: A(x) = 2tan⁻¹(x) [using identity: if x = tan(θ), then 2x/(1+x²) = sin(2θ)]<br>- And: B(x) is an odd function [tan⁻¹(-u) = -tan⁻¹(u)]</p><p><strong>Step 3:</strong> Write I = ∫[2tan⁻¹(x) + B(x)]/(e^x + 1) dx over [-1/√3, 1/√3]</p><p><strong>Step 4:</strong> Since B(x) is odd and (e^x + 1)⁻¹ is neither even nor odd, use:<br>I = ∫ 2tan⁻¹(x)/(e^x + 1) dx + ∫ B(x)/(e^x + 1) dx<br>The second integral = 0 (odd function property applies when denominator is asymmetric)</p><p><strong>Step 5:</strong> For the remaining integral, apply: ∫ 2tan⁻¹(x)/(e^x + 1) dx from -1/√3 to 1/√3<br>Using f(x) + f(-x) technique with 1/(e^x + 1) + 1/(e^(-x) + 1) = 1:<br>I = ∫ 2tan⁻¹(x) · [1/2] dx = ∫ tan⁻¹(x) dx over the symmetric interval = 0 (net cancellation)</p><p>∴ Answer: <strong>A (which is 0)</strong></p>
Correct Answer: A