Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>If \(2\displaystyle\int_0^1 \tan^{-1}x\,dx = \int_0^1 \cot^{-1}(1-x+x^2)\,dx\), then \(\displaystyle\int_0^1 \tan^{-1}(1-x+x^2)\,dx\) is equal to</p>
<p>\(\dfrac{\pi}{2} + \log 2\)</p>
<p>\(\log 2\)</p>
<p>\(\dfrac{\pi}{2} - \log 4\)</p>
<p>\(\log 4\)</p>
Step-by-Step Solution
Key Concept: Use the complementary inverse trigonometric identity tan⁻¹(y) + cot⁻¹(y) = π/2, combined with substitution and the given condition to relate the three integrals systematically.
<p><strong>Step 1:</strong> Use the fundamental identity for inverse functions:</p><p>tan⁻¹(y) + cot⁻¹(y) = π/2</p><p>Therefore: tan⁻¹(1-x+x²) + cot⁻¹(1-x+x²) = π/2</p><p><strong>Step 2:</strong> Integrate both sides from 0 to 1:</p><p>∫₀¹ tan⁻¹(1-x+x²)dx + ∫₀¹ cot⁻¹(1-x+x²)dx = π/2</p><p><strong>Step 3:</strong> From the given condition:</p><p>2∫₀¹ tan⁻¹(x)dx = ∫₀¹ cot⁻¹(1-x+x²)dx</p><p><strong>Step 4:</strong> Substitute this into the equation from Step 2:</p><p>∫₀¹ tan⁻¹(1-x+x²)dx + 2∫₀¹ tan⁻¹(x)dx = π/2</p><p><strong>Step 5:</strong> Calculate ∫₀¹ tan⁻¹(x)dx using integration by parts:</p><p>∫₀¹ tan⁻¹(x)dx = [x·tan⁻¹(x)]₀¹ - ∫₀¹ x/(1+x²)dx = π/4 - (1/2)ln(2)</p><p><strong>Step 6:</strong> Substitute back:</p><p>∫₀¹ tan⁻¹(1-x+x²)dx = π/2 - 2[π/4 - (1/2)ln(2)]</p><p>∫₀¹ tan⁻¹(1-x+x²)dx = π/2 - π/2 + ln(2)</p><p>∴ Answer: <strong>ln(2)</strong></p>
Correct Answer: C