Let fₙ=∫0^(π/2)(Σk=1ⁿ sin ᵏ⁻¹ x)(Σk=1ⁿ(2 k-1) sin ᵏ⁻¹ x) cos x d x, n ∈ ℕ. Then f₂₁-f₂₀ is equal to _________

Solution

Maths sir · 4 steps

Put sin x = y² and the integrand becomes 2PP', the derivative of a perfect square.

1. First substitution. Put t = sin x, so dt = cos x dx and the limits become 0 to 1. Then fₙ = ∫₀¹ (Σₖ₌₁ⁿ tᵏ⁻¹)(Σₖ₌₁ⁿ(2k-1)tᵏ⁻¹)dt.
2. Second substitution. The coefficients 1, 3, 5, … come from differentiating odd powers, so put t = y², dt = 2y dy, limits still 0 to 1. Define P(y) = y + y³ + … + y²ⁿ⁻¹ = Σₖ₌₁ⁿ y²ᵏ⁻¹.
3. Spot the derivative. Now yΣₖ₌₁ⁿ y²ᵏ⁻² = P(y) and Σₖ₌₁ⁿ(2k-1)y²ᵏ⁻² = P'(y). So the whole integrand is 2P(y)P'(y), which is the derivative of P(y)².

   fₙ = ∫₀¹ 2P(y)P'(y) dy = [P(y)²]₀¹ = (1+1+…+1ₙ)² = n²

4. So. f₂₁ - f₂₀ = 21² - 20² = 441 - 400 = 41.

The idea to keep. When one sum has coefficients 1, 3, 5, …, read it as the derivative of odd powers. Choose the substitution that turns the integrand into P P', and the integral is just 1/2P².

Answer from the NTA final answer key. Solution by PrepWiser.