If ∫₀^π (sin ³x)e^(- sin ²x)dx = α - β/e∫₀¹ √t eᵗdt , then α + β is equal to ____________.

Solution

Maths sir · 4 steps

Peel off one sin x, put cos x = u, then t = u² turns it into the integral they gave you.

1. Substitute cos x. Write sin³ x = (1-cos² x)sin x and e^(-sin² x) = e^(cos² x - 1), then put u = cos x, so du = -sin x dx and the limits go from 1 to -1. This gives ∫₋₁¹(1-u²)e^(u²-1) du. The integrand is even, so it equals 2/e∫₀¹ (1-u²)e^(u²) du.
2. Second substitution. With t = u² we have du = dt/2√t and the limits stay 0 to 1. The integral becomes 1/e∫₀¹ (t^(-1/2) - t^(1/2))eᵗ dt.
3. Parts on first piece. Call K = ∫₀¹ √t eᵗ dt. By parts, ∫₀¹ t^(-1/2)eᵗ dt = [2√t eᵗ]₀¹ - 2∫₀¹ √t eᵗ dt = 2e - 2K.

   ∫₀^π sin³ x e^(-sin² x) dx = 1/e(2e - 2K - K) = 2 - 3/e∫₀¹ √t eᵗ dt

4. So. Matching with α - β/eK gives α = 2 and β = 3, so α + β = 5.

The common slip: 2.5. Forgetting to double the integral when folding ∫₋₁¹ of the even integrand onto [0,1] gives α = 1 and β = 3/2, so the sum comes out as 2.5, not 5.

The idea to keep. When the answer is written in terms of an integral you cannot evaluate, do not try to evaluate it. Reshape your integral by substitution and by parts until that same integral appears.

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