A-LevelFurther MathematicsIntegration9231/12 · May/June 2018 · Q910 Marks

It is given that, for n ≥ 0, Iₙ = ∫⁰¹ᐟ⁴π secⁿx tan²x dx.

(i)Using the substitution u = tan x, or otherwise, find ∫sec²x tan²x dx.[2]
(ii)Using the result that d/dx (sec x) = tan x sec x, show that, for n ≥ 2, (n + 1)Iₙ = (√2)ⁿ⁻² + (n − 2)Iₙ₋₂.[5]
(iii)Hence find the mean value of sec⁴x tan²x with respect to x over the interval 0 ≤ x ≤ ¼π, giving your answer in exact form.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

M1 (i) Uses substitution correctly. du/dx = sec²x so integral becomes ∫u² du
A1 (i) 1/3 tan³x + C
B1 (ii) Separates into correct structure. Iₙ = ∫ secⁿ⁻²x sec²x tan²x dx
M1 (ii) Uses integration by parts correctly. = [1/3 secⁿ⁻²x tan³x]₀¹ᐟ⁴π - (n-2)/3 ∫ secⁿ⁻²x tan⁴x dx
M1 (ii) Uses tan²x = sec²x-1 to rewrite tan⁴x. ∫ secⁿ⁻²x tan⁴x dx = ∫ secⁿ⁻²x tan²x (sec²x-1) dx = ∫ secⁿx tan²x dx - ∫ secⁿ⁻²x tan²x dx = I_n - I_(n-2).
M1 (ii) Steps for full derivation required for M1
A1 (ii) (n+1)Iₙ = (√2)ⁿ⁻² + (n − 2)Iₙ₋₂. (AG)
B1 FT (iii) Attempts to find I₂. From (n+1)Iₙ = (√2)ⁿ⁻² + (n − 2)Iₙ₋₂, for n=2: 3I₂ = (√2)⁰ + 0 = 1, so I₂ = 1/3.
M1 (iii) Uses reduction formula. Mean value = (1/(π/4)) I₄. For n=4: 5I₄ = (√2)² + 2I₂ = 2 + 2(1/3) = 8/3, so I₄ = 8/15.
A1 (iii) Mean value = (4/π) (8/15) = 32/(15π).

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