A civil engineer is designing a curved water channel whose cross-sectional area is modelled by the definite integral A = ∫₀^π x · sin x / (1 + cos²x) dx, where x is measured in metres. The engineer uses properties of definite integrals to evaluate this integral analytically.
A civil engineer is designing a curved water channel whose cross-sectional area is modelled by the definite integral
A = ∫₀^π x · sin x / (1 + cos²x) dx
where x is measured in metres.
(i) Using the property ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a+b−x) dx, show that A = (π/2) ∫₀^π sin x / (1 + cos²x) dx. [2]
(ii) Hence evaluate A and state the cross-sectional area of the channel in sq. metres. [2]
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Let I = ∫₀^π x sin x / (1 + cos²x) dx … (i)
Applying King's property ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx with a = 0, b = π,
I = ∫₀^π (π − x) sin(π − x) / (1 + cos²(π − x)) dx
∵ sin(π − x) = sin x and cos(π − x) = −cos x, so cos²(π − x) = cos²x,
⟹ I = ∫₀^π (π − x) sin x / (1 + cos²x) dx … (ii)
Adding (i) and (ii),
2I = ∫₀^π [x sin x + (π − x) sin x] / (1 + cos²x) dx
⟹ 2I = π ∫₀^π sin x / (1 + cos²x) dx
∴ I = (π/2) ∫₀^π sin x / (1 + cos²x) dx Hence shown.
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Part (ii) — [2 marks]
Let J = ∫₀^π sin x / (1 + cos²x) dx
Put t = cos x ⟹ dt = −sin x dx.
When x = 0, t = 1; when x = π, t = −1.
J = ∫₁^(−1) (−dt) / (1 + t²) = ∫₋₁^1 dt / (1 + t²)
⟹ J = [tan⁻¹t]₋₁¹ = tan⁻¹(1) − tan⁻¹(−1)
⟹ J = π/4 − (−π/4) = π/2
∴ A = I = (π/2) · J = (π/2) · (π/2) = π²/4
∴ The cross-sectional area of the channel is π²/4 sq. metres.