If $f(x) = x + \int_1^x\frac{f(t)}{t}dt$, then $\int_0^{\pi}\frac{(f(\sin\theta) - \sin\theta)}{\sin\theta}d\theta$ is equal to:
Step-by-Step Solution
Key Concept: Differentiate the integral equation to get f'(x) = 1 + f(x)/x, solve the resulting first-order linear ODE using integrating factor 1/x to obtain f(x) = x(ln x + 1), then evaluate the definite integral ∫₀^π [(f(sin θ) - sin θ)/sin θ] dθ = ∫₀^π ln(sin θ) dθ = -π ln 2.
Given $f'(x) = 1 + rac{f(x)}{x}$, rewrite as $ rac{dy}{dx} - rac{y}{x} = 1$. The integrating factor is $e^{-\int rac{1}{x}dx} = rac{1}{x}$. Multiplying through: $y \cdot rac{1}{x} = \int 1 \cdot rac{1}{x}dx + c = \ln x + c$, so $y = x\ln x + cx$. Using $c = 1$ from boundary conditions, $ rac{f(x) - x}{x} = \ln x \Rightarrow f(x) = \int_0^{\pi}\ln(\sin heta)d heta = -\pi\ln 2$.
Correct Answer: 3