Evaluate ∫₀^(π/2) x sin x dx. Given that I_n = ∫₀^(π/2) xⁿ sin x dx, prove that, for n > 1, I_n = n(π/2)ⁿ⁻¹ – n(n - 1)I_n₋₂. By first using the substitution x = cos⁻¹u, find the value of ∫₀¹ (cos⁻¹u)³ du, giving your answer in an exact form.

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Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.

1. [M1A1] ∫₀^(π/2) x sin x dx. Integration by parts: [-x cos x]₀^(π/2) + ∫₀^(π/2) cos x dx = (0 - 0) + [sin x]₀^(π/2) = 1.
2. [M1A1] For I_n = ∫₀^(π/2) xⁿ sin x dx, integrate by parts: [-xⁿ cos x]₀^(π/2) + ∫₀^(π/2) nxⁿ⁻¹ cos x dx. The first term is 0 at both limits.
3. [A1] Apply integration by parts again to ∫₀^(π/2) nxⁿ⁻¹ cos x dx: [nxⁿ⁻¹ sin x]₀^(π/2) - ∫₀^(π/2) n(n-1)xⁿ⁻² sin x dx. The first term evaluates to n(π/2)ⁿ⁻¹ (at π/2) - 0 (at 0).
4. [A1] Resulting in I_n = n(π/2)ⁿ⁻¹ – n(n-1)I_n₋₂.
5. [B1B1] Substitution x = cos⁻¹u => u = cos x. du = -sin x dx. When u=0, x=π/2. When u=1, x=0. So ∫₀¹ (cos⁻¹u)³ du = ∫_π/2⁰ x³ (-sin x) dx.
6. [B1] = ∫₀^(π/2) x³ sin x dx = I₃.
7. [M1A1] Use reduction formula for I₃: I₃ = 3(π/2)² - 3(2)I₁. I₁ = ∫₀^(π/2) x sin x dx = 1 (from first part). I₃ = 3(π²/4) - 6(1) = 3π²/4 - 6.