A-Level Techniques of integration: worked solution

Question

Hence evaluate ∫₀^{π/2} |2x-1| cos x dx, giving the answer as A-4cos B with A, B exact constants.

A modulus cannot be integrated directly, so first split the range at the point where 2x − 1 changes sign. 2x − 1 = 0 gives x = 1/2, and 0 < 1/2 < π/2, so the split point lies inside the range of integration. For 0 ≤ x < 1/2: 2x − 1 < 0, so |2x − 1| = −(2x − 1) = 1 − 2x For 1/2 ≤ x ≤ π/2: 2x − 1 ≥ 0, so |2x − 1| = 2x − 1 Hence ∫₀^{π/2} |2x−1| cos x dx = −∫₀^{1/2} (2x−1) cos x dx + ∫_{1/2}^{π/2} (2x−1) cos x dx Both pieces need the same antiderivative, so find ∫ (2x−1) cos x dx by parts, differentiating the polynomial (which reduces it to a constant) and integrating the cosine: u = 2x − 1, dv/dx = cos x, so du/dx = 2 and v = sin x ∫ (2x−1) cos x dx = (2x−1) sin x − ∫ 2 sin x dx = (2x−1) sin x − (−2 cos x) = (2x−1) sin x + 2 cos x Call this F(x) = (2x − 1) sin x + 2 cos x and evaluate it at the three x-values needed (x in RADIANS): F(0) = (2(0) − 1) sin 0 + 2 cos 0 = (−1)(0) + 2(1) = 2 F(1/2) = (2(1/2) − 1) sin(0.5) + 2 cos(0.5) = (0) sin(0.5) + 2 cos(0.5) = 2 cos 0.5 F(π/2) = (π − 1) sin(π/2) + 2 cos(π/2) = (π − 1)(1) + 2(0) = π − 1 First piece: −[ F(1/2) − F(0) ] = −(2 cos 0.5 − 2) = 2 − 2 cos 0.5 Second piece: F(π/2) − F(1/2) = (π − 1) − 2 cos 0.5 Add them: (2 − 2 cos 0.5) + (π − 1 − 2 cos 0.5) = π + 1 − 4 cos 0.5 This is already in the form A − 4 cos B, so A = π + 1 and B = 0.5 (that is, 1/2, in radians).