Let f(x)=∫0ˣ g(t) log ₑ((1-t)/(1+t)) dt, where g is a continuous odd function. If ∫_(-π / 2)^(π / 2)(f(x)+(x² cos x)/(1+eˣ)) d x=(π/α)²-α, then α is equal to _________.

Solution

Maths sir · 4 steps

Split the integral: f is odd, so it vanishes, and the 1/(1+eˣ) factor drops out by pairing.

1. f is odd. logₑ(1-t)/(1+t) is odd, since replacing t by -t flips the fraction. Odd g times an odd function is even, and ∫₀ˣ of an even function is odd. So ∫_(-π/2)^(π/2) f(x) dx = 0.
2. Pair opposite points. Let h(x) = (x²cos x)/(1+eˣ). Then h(-x) = (eˣ x²cos x)/(1+eˣ), so h(x)+h(-x) = x²cos x. That gives ∫₋ₐᵃ h dx = ∫₀ᵃ(h(x)+h(-x))dx.

   ∫_(-π/2)^(π/2)(x²cos x)/(1+eˣ) dx = ∫₀^(π/2) x²cos x dx = [x²sin x + 2x cos x - 2 sin x]₀^(π/2)

3. Evaluate. At π/2 the bracket is π²/4 + 0 - 2, and at 0 it is 0. So the whole integral is π²/4 - 2 = (π/2)² - 2.
4. So. Matching with (π/α)² - α gives α = 2.

The idea to keep. On [-a, a], pair x with -x. Odd parts vanish, and a factor like 1/(1+eˣ) always collapses because 1/(1+eˣ)+eˣ/(1+eˣ) = 1.

Answer from the NTA final answer key. Solution by PrepWiser.