Let f(x) = sin x and g(x) - f(x) = f(4-x). What is ∫₀⁴ f(4-x)/g(4-x) dx equal to?

Since g(4-x) = f(4-x) + f(x) = g(x), the integral ∫₀⁴ f(4-x)/g(4-x) dx = ∫₀⁴ f(4-x)/g(x) dx. Using substitution u = 4-x, this equals ∫₀⁴ f(u)/g(4-u) du = ∫₀⁴ f(u)/g(u) du = 2 (from previous problem). Hence the value is 2.