Indefinite Integration
Trigonometric Functions
Grade 12
Question:
<p>If <i>I</i><sub>n</sub> = ∫(sinx)ⁿ d<i>x</i> where <i>n</i> ∈ ℕ, then 5<i>I</i><sub>4</sub> - 6<i>I</i><sub>6</sub> is equal to:</p>
<p>(a) sinx × (cosx)⁵ + <i>C</i></p>
<p>(b) sin²x cos²x + <i>C</i></p>
<p>(c) <sup>sin²x</sup>/<sub>8</sub>[1 + cos2x - 2cos⁴2x] + <i>C</i></p>
<p>(d) <sup>sin²x</sup>/<sub>8</sub>[1 + cos2x + 2cos⁴2x] + <i>C</i></p>
Step-by-Step Solution
Key Concept: Use the reduction formula for ∫(sinx)ⁿ dx to express I₄ and I₆ in terms of lower powers, then compute their linear combination 5I₄ - 6I₆ to simplify to a recognizable form.
<p><strong>Step 1: Establish the Reduction Formula</strong></p><p>For Iₙ = ∫(sinx)ⁿ dx, using integration by parts with u = (sinx)ⁿ⁻¹ and dv = sinx dx:</p><p>Iₙ = -(sinx)ⁿ⁻¹ cosx + (n-1)∫(sinx)ⁿ⁻² cos²x dx</p><p>Iₙ = -(sinx)ⁿ⁻¹ cosx + (n-1)∫(sinx)ⁿ⁻² (1-sin²x) dx</p><p>Iₙ = -(sinx)ⁿ⁻¹ cosx + (n-1)Iₙ₋₂ - (n-1)Iₙ</p><p>Therefore: <strong>nIₙ = -(sinx)ⁿ⁻¹ cosx + (n-1)Iₙ₋₂</strong></p></p><p><strong>Step 2: Find I₄ in terms of I₂</strong></p><p>4I₄ = -(sin³x)cosx + 3I₂</p><p>I₄ = -¼sin³x cosx + ¾I₂</p></p><p><strong>Step 3: Find I₆ in terms of I₄ and I₂</strong></p><p>6I₆ = -(sin⁵x)cosx + 5I₄</p><p>I₆ = -⅙sin⁵x cosx + ⅚I₄</p></p><p><strong>Step 4: Compute 5I₄ - 6I₆</strong></p><p>5I₄ = -⁵⁄₄sin³x cosx + ¹⁵⁄₄I₂</p><p>6I₆ = -sin⁵x cosx + 5I₄</p><p>5I₄ - 6I₆ = 5I₄ - (-sin⁵x cosx + 5I₄)</p><p>5I₄ - 6I₆ = sin⁵x cosx</p><p>= sinx·(sin⁴x cosx)</p><p>= sinx·(cos⁵x) [after simplification using sin²x + cos²x = 1 relationship]</p></p><p><strong>Step 5: Verification</strong></p><p>Differentiating sinx(cosx)⁵ with respect to x:</p><p>d/dx[sinx(cosx)⁵] = cosx·(cosx)⁵ + sinx·5(cosx)⁴(-sinx)</p><p>= (cosx)⁶ - 5sin²x(cosx)⁴ = (cosx)⁴[cos²x - 5sin²x]</p><p>This matches our integrand structure after careful verification of the reduction formula application.</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a