Definite Integration
Reduction formulae for definite integrals
Grade 12
Question:
<p>If \( I_n = \displaystyle\int_0^{\pi/2} \dfrac{\sin^2 nx}{\sin x} \, dx \), then \( I_{n+1} + I_{n-1} - 2I_n \) equals:</p>
<p>(a) \( 0 \)</p>
<p>(b) \( \dfrac{1}{2n} \)</p>
<p>(c) \( \dfrac{1}{n} \)</p>
<p>(d) \( -\dfrac{1}{2n} \)</p>
Step-by-Step Solution
Key Concept: Use the recurrence relation derived from the Chebyshev polynomial identity: sin(2nx) = 2sin(nx)cos(nx). This creates a telescoping structure when you compute differences of consecutive I_n values.
<p><strong>Step 1:</strong> Write the expression for I_{n+1} + I_{n-1} - 2I_n</p><p>I_{n+1} + I_{n-1} - 2I_n = ∫₀^(π/2) [sin²((n+1)x) + sin²((n-1)x) - 2sin²(nx)]/sin(x) dx</p><p><strong>Step 2:</strong> Apply the identity: sin²A - sin²B = sin(A+B)sin(A-B)</p><p>sin²((n+1)x) - sin²(nx) = sin((2n+1)x)sin(x)</p><p>sin²((n-1)x) - sin²(nx) = sin((2n-1)x)·(-sin(x)) = -sin((2n-1)x)sin(x)</p><p><strong>Step 3:</strong> Combine to get:</p><p>sin²((n+1)x) + sin²((n-1)x) - 2sin²(nx) = sin(x)[sin((2n+1)x) - sin((2n-1)x)]</p><p><strong>Step 4:</strong> Apply sum-to-product: sin A - sin B = 2cos((A+B)/2)sin((A-B)/2)</p><p>sin((2n+1)x) - sin((2n-1)x) = 2cos(2nx)sin(x)</p><p><strong>Step 5:</strong> Therefore:</p><p>I_{n+1} + I_{n-1} - 2I_n = ∫₀^(π/2) sin(x)·2cos(2nx)sin(x)/sin(x) dx = 2∫₀^(π/2) cos(2nx) dx</p><p><strong>Step 6:</strong> Evaluate: 2∫₀^(π/2) cos(2nx) dx = 2[sin(2nx)/(2n)]₀^(π/2) = [sin(nπ)]/n = 0 (for integer n)</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: A