The value of ∫_(- π /2)^(π /2) (cos ²x)/(1 + 3ˣ) dx is :

Solution

Maths sir · 4 steps

Replace x by -x and add the two integrals: the awkward 3ˣ cancels out completely.

1. Flip the variable. Call the integral I. The limits are symmetric, so ∫₋ₐᵃ f(x) dx = ∫₋ₐᵃ f(-x) dx. Since cos²(-x) = cos² x and 1/(1+3⁻ˣ) = 3ˣ/(3ˣ+1), this gives I = ∫_(-π/2)^(π/2) (3ˣ cos² x)/(1+3ˣ) dx.
2. Add both forms. Add the original form and the new form. The fractions combine to (1+3ˣ)/(1+3ˣ) = 1, so only cos² x is left.

   2I = ∫_(-π/2)^(π/2) cos² x dx = ∫_(-π/2)^(π/2) (1+cos 2x)/2 dx = π/2

3. Evaluate. The 1/2 part gives 1/2 × π = π/2. The (cos 2x)/2 part gives (sin 2x)/4, which is 0 at both x = π/2 and x = -π/2.
4. So. 2I = π/2, so I = π/4, which is option (d).

The idea to keep. When you see 1/(1+aˣ) with symmetric limits, use x → -x and add. The two fractions sum to 1, and the answer is half of the integral of the remaining even function.

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