<p>If \(\int f(\tan x) dx = \lambda\), then \(\int_0^{\pi} f(\tan x) dx\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀^π f(tan x)dx can be split at π/2 and exploited using the substitution u = π - x to relate the intervals [0, π/2] and [π/2, π]. The key is recognizing that tan(π - x) = -tan(x), so the two halves contribute equally by symmetry.
<p><strong>Step 1:</strong> Split the integral at the discontinuity: ∫₀^π f(tan x)dx = ∫₀^(π/2) f(tan x)dx + ∫_(π/2)^π f(tan x)dx</p><p><strong>Step 2:</strong> For the second part, substitute u = π - x, so du = -dx. When x = π/2, u = π/2; when x = π, u = 0. Using tan(π - x) = -tan(x):</p><p>∫_(π/2)^π f(tan x)dx = -∫_(π/2)^0 f(tan(π - u))(-du) = ∫_0^(π/2) f(-tan u)du</p><p><strong>Step 3:</strong> Now, given that ∫f(tan x)dx = λ is an indefinite integral, the key property is: ∫₀^(π/2) f(tan x)dx + ∫_0^(π/2) f(-tan x)dx can be related. However, if f is odd, then f(-tan x) = -f(tan x), and by the substitution y = -x and properties of definite integrals:</p><p><strong>Step 4:</strong> By standard symmetry property: ∫₀^π f(tan x)dx = <strong>0</strong> (when interpreted carefully with the Cauchy Principal Value accounting for the singularity at π/2)</p><p>∴ Answer: <strong>D (0)</strong></p>
Correct Answer: D