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∫₀^π x f(sin x) dx is equal to

1. π ∫₀^π f(cos x) dx
2. π ∫₀^π f(sin x) dx
3. (π/2) ∫₀^(π/2) f(sin x) dx
4. π ∫₀^(π/2) f(cos x) dx

Correct answer: π ∫₀^(π/2) f(cos x) dx

Solution

By x -> pi - x, ∫0^pi x f(sin x)dx = (pi/2)∫0^pi f(sin x)dx = pi∫0^(pi/2) f(sin x)dx. Since ∫0^(pi/2) f(sin x)dx = ∫0^(pi/2) f(cos x)dx, this equals pi∫0^(pi/2) f(cos x)dx.

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