Definite Integration
Reduction formulae for definite integrals
Grade 12
Question:
<p>Let \( I_n = \displaystyle\int_0^{\pi/2} \dfrac{\sin^2 nx}{\sin x} \, dx \) and \( I_{n+1} = \displaystyle\int_0^{\pi/2} \dfrac{\sin^2(n+1)x}{\sin x} \, dx \). Then \( I_{n+1} - I_n \) equals:</p>
<p>(a) \( \dfrac{1}{2n+1} \)</p>
<p>(b) \( \dfrac{1}{2n-1} \)</p>
<p>(c) \( \dfrac{2}{2n+1} \)</p>
<p>(d) \( 0 \)</p>
Step-by-Step Solution
Key Concept: Use the identity sin²(n+1)x - sin²(nx) = sin(2n+1)x·sin(x) to convert the difference of integrals into a single integral, which simplifies dramatically due to the sin(x) denominator cancellation.
<p><strong>Step 1: Set up the difference</strong></p><p>I_{n+1} - I_n = ∫₀^(π/2) [sin²((n+1)x) - sin²(nx)]/sin(x) dx</p><p><strong>Step 2: Factor the numerator using algebraic identity</strong></p><p>sin²A - sin²B = (sinA - sinB)(sinA + sinB)</p><p>sin²((n+1)x) - sin²(nx) = [sin((n+1)x) - sin(nx)][sin((n+1)x) + sin(nx)]</p><p><strong>Step 3: Apply sum-to-product formulas</strong></p><p>sin((n+1)x) - sin(nx) = 2cos((2n+1)x/2)sin(x/2)</p><p>sin((n+1)x) + sin(nx) = 2sin((2n+1)x/2)cos(x/2)</p><p><strong>Step 4: Substitute and simplify</strong></p><p>Numerator = 2cos((2n+1)x/2)sin(x/2) · 2sin((2n+1)x/2)cos(x/2)</p><p>= 4sin((2n+1)x/2)cos((2n+1)x/2)sin(x/2)cos(x/2)</p><p>= 2sin((2n+1)x)·(1/2)sin(x) [using 2sinθcosθ = sin(2θ)]</p><p>= sin((2n+1)x)sin(x)</p><p><strong>Step 5: Cancel and integrate</strong></p><p>I_{n+1} - I_n = ∫₀^(π/2) [sin((2n+1)x)sin(x)]/sin(x) dx = ∫₀^(π/2) sin((2n+1)x) dx</p><p>= [-cos((2n+1)x)/(2n+1)]₀^(π/2)</p><p>= -[cos((2n+1)π/2) - cos(0)]/(2n+1)</p><p>= -[0 - 1]/(2n+1) = <strong>1/(2n+1)</strong></p><p>∴ Answer: A</p>
Correct Answer: A