A-Level Techniques of integration: worked solution

Question

Using the previous result, or by another method, find the exact value of ∫₀^{π/6} sin2θ cos3θ dθ.

The previous part supplies the triple-angle identity cos3θ = 4cos³θ - 3cosθ. Combined with the double-angle identity for sin2θ, it turns the product into terms of the form sinθ × (power of cosθ), each of which integrates by inspection. Step 1: replace both factors by expressions in sinθ and cosθ. sin2θ = 2 sinθ cosθ cos3θ = 4cos³θ - 3cosθ so ∫₀^{π/6} sin2θ cos3θ dθ = ∫₀^{π/6} (2 sinθ cosθ)(4cos³θ - 3cosθ) dθ Step 2: multiply out the bracket, adding the powers of cosθ. (2 sinθ cosθ)(4cos³θ) = 8 sinθ cos⁴θ (2 sinθ cosθ)(-3cosθ) = -6 sinθ cos²θ so the integrand is 8 sinθ cos⁴θ - 6 sinθ cos²θ Step 3: integrate each term. Since d/dθ (cos^(n+1)θ) = -(n + 1)cosⁿθ sinθ, reversing this gives the standard result ∫ sinθ cosⁿθ dθ = -cos^(n+1)θ/(n + 1) Applying it with n = 4 and n = 2: ∫ 8 sinθ cos⁴θ dθ = 8 × (-cos⁵θ/5) = -8cos⁵θ/5 ∫ -6 sinθ cos²θ dθ = -6 × (-cos³θ/3) = 2cos³θ So the antiderivative is -8cos⁵θ/5 + 2cos³θ Step 4: substitute the upper limit θ = π/6, where cosθ = √3/2. cos³θ = (√3/2)³ = 3√3/8 cos⁵θ = (√3/2)⁵ = 9√3/32 Value = -(8/5)(9√3/32) + 2(3√3/8) = -72√3/160 + 6√3/8 = -9√3/20 + 3√3/4 = -9√3/20 + 15√3/20 = 6√3/20 = 3√3/10 Step 5: substitute the lower limit θ = 0, where cosθ = 1. Value = -(8/5)(1) + 2(1) = -8/5 + 10/5 = 2/5 Step 6: subtract the lower value from the upper value. ∫₀^{π/6} sin2θ cos3θ dθ = 3√3/10 - 2/5