The value of ∫_((- π)/2)^(π/2) 1/(1 + eˢⁱⁿˣ)dx is:

Solution

Maths sir · 3 steps

Replace x by -x, add the two forms, and the ugly exponential turns into 1.

1. Use the property. For symmetric limits, ∫₋ₐᵃ f(x) dx = ∫₋ₐᵃ f(-x) dx. Call the integral I and replace x by -x. Since sin(-x) = -sin x, the integrand becomes 1/(1+e⁻ˢⁱⁿˣ).
2. Tidy the new form. Multiply top and bottom by eˢⁱⁿˣ: 1/(1+e⁻ˢⁱⁿˣ) = eˢⁱⁿˣ/(eˢⁱⁿˣ+1). So I = ∫_(-π/2)^(π/2) eˢⁱⁿˣ/(1+eˢⁱⁿˣ) dx.

   2I = ∫_(-π/2)^(π/2) (1 + eˢⁱⁿˣ)/(1+eˢⁱⁿˣ) dx = ∫_(-π/2)^(π/2) 1 dx = π

3. So. Adding the two forms of I gives 2I = π, so I = π/2, which is option (c).

The idea to keep. When the limits are -a to a and the integrand looks like 1/(1+eᵍ⁽ˣ⁾) with g odd, add f(x) and f(-x). They sum to 1, so the answer is half the interval length.

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