It is given that I_n = ∫(from 0 to π/4) (sin^(2n+2) x) / (cos x) dx, where n ≥ 0. Show that I_n – I_(n+1) = (2^(−(n+2))) / (2n + 1). [5] Hence show that ∫(from 0 to π/4) (sin⁶ x) / (cos x) dx = ln(1 + √2) – 73√2 / 120. [5]

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

1. [B1] I_n+1 = ∫(from 0 to π/4) (sin^(2n+2) x) / (cos x) dx (LR)
2. [M1] = ∫(from 0 to π/4) ( (1-cos²x)sin^(2n) x ) / (cos x) dx (LR)
3. [A1] = I_n - ∫(from 0 to π/4) cos x sin^(2n) x dx
4. [M1] = I_n - [ sin^(2n+1)x / (2n+1) ] (from 0 to π/4)
5. [A1] ⇒ I_n – I_(n+1) = (2^-(n+2))/(2n+1) (AG)
6. [M1A1] I_0 = ∫(from 0 to π/4) sec x dx = [ln|sec x + tan x|] (from 0 to π/4) = ln(1+√2)
7. [M1] I_1 = ln(1+√2) - 1/√2
8. [A1] I_2 = ln(1+√2) - 1/√2 - 1/(6√2)
9. [A1] I_3 = ln(1+√2) - 1/√2 - 1/(6√2) - 1/(20√2) = ln(1+√2) - 73√2/120 (AG)