A-LevelFurther MathematicsIntegration9231/12 · May/June 2011 · Q1012 Marks

Let Iₙ = ∫₀^(π/2) cosⁿ x dx , where n ≥ 0 .

Show that, for all n ≥ 2 , Iₙ = (n-1)/n Iₙ₋₂ .[4]
A curve has parametric equations x = a sin³ t and y = a cos³ t , where a is a constant and 0 ≤ t ≤ π/2 . Show that the mean value m of y over the interval 0 ≤ x ≤ a is given by m = 3a ∫₀^(π/2) (cos⁴ t - cos⁶ t) dt .[4]
Find the exact value of m, in terms of a.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.

[M1A1] Puts cosⁿ x = cosⁿ⁻¹ x · cos x and integrates by parts. Iₙ = [cosⁿ⁻¹ x sin x]₀^(π/2) + (n-1)∫₀^(π/2) cosⁿ⁻² x sin² x dx
[M1] Uses cos² x = 1-sin² x and obtains reduction formula. Iₙ = (n-1) ∫₀^(π/2) (cosⁿ⁻² x - cosⁿ x) dx
[A1] ⇒ Iₙ = (n-1)/n Iₙ₋₂ (n ≥ 2 ) (AG)
[M1] Uses mean value formula. Mean value = (∫ y dx)/(b-a)
[M1A1] Transforms variable to t. 1/a ∫₀^(π/2) a cos³ t · 3a sin² t cos t dt
[A1] Uses sin² t = 1-cos² t to write in a form where reduction formula can be used. = 3a ∫₀^(π/2) (cos⁴ t - cos⁶ t) dt (AG)
[B1] Finds I₀ . I₀ = π/2 or I₂ = π/4
[M1A1] Uses reduction formula to find I₄ and I₆ . I₄ = 3/4 I₂ = 3/4 · π/4 = 3π/16 , I₆ = 5/6 I₄ = 5/6 · 3π/16 = 5π/32
[A1] Substitutes in integral to find mean value. Mean value = 3aπ/32

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