Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>\(\int_0^{\pi} x f(\sin x)\, dx\) is equal to</p>
<p>\(\pi \int_0^{\pi} f(\cos x)\, dx\)</p>
<p>\(\pi \int_0^{\pi} f(\sin x)\, dx\)</p>
<p>\(\frac{\pi}{2} \int_0^{\pi/2} f(\sin x)\, dx\)</p>
<p>\(\pi \int_0^{\pi/2} f(\cos x)\, dx\)</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx by substituting x → π - x and exploiting the symmetry that sin(π - x) = sin x.
<p><strong>Step 1:</strong> Let I = ∫₀^π x f(sin x) dx</p><p><strong>Step 2:</strong> Substitute x → (π - x), so dx → -dx. The limits transform from [0, π] to [π, 0]:<br/>I = ∫^0_π (π - x) f(sin(π - x)) (-dx) = ∫₀^π (π - x) f(sin x) dx</p><p><strong>Step 3:</strong> Since sin(π - x) = sin x, we have:<br/>I = ∫₀^π (π - x) f(sin x) dx = π∫₀^π f(sin x) dx - ∫₀^π x f(sin x) dx = π∫₀^π f(sin x) dx - I</p><p><strong>Step 4:</strong> Solving for I:<br/>2I = π∫₀^π f(sin x) dx<br/>I = (π/2)∫₀^π f(sin x) dx</p><p>∴ Answer: D</p>
Correct Answer: D