The value of the definite integral
∫_(- π/4)^(π/4) dx/(1 + eˣᶜᵒˢˣ)(sin ⁴x + cos ⁴x) is equal to :

Solution

Maths sir · 4 steps

Replace x by -x and add: the ugly exponential factor adds up to exactly 1 and disappears.

1. Replace x by -x. Call the integral I. Since ∫₋ₐᵃ f(x) dx = ∫₋ₐᵃ f(-x) dx, replace x by -x. Here x cos x is odd, so it becomes -x cos x, while sin⁴ x + cos⁴ x does not change.
2. Add the two forms. Let g = x cos x. Then 1/(1+eᵍ) + 1/(1+e⁻ᵍ) = 1/(1+eᵍ) + eᵍ/(eᵍ+1) = 1. So 2I = ∫_(-π/4)^(π/4) dx/(sin⁴ x + cos⁴ x). This integrand is even, so I = ∫₀^(π/4) dx/(sin⁴ x + cos⁴ x).
3. Tangent substitution. Divide the top and bottom by cos⁴ x to get (sec⁴ x)/(1+tan⁴ x). With t = tan x and dt = sec² x dx, this becomes ∫₀¹ (1+t²)/(1+t⁴) dt. Divide the top and bottom by t² and put u = t - 1/t. As t goes from 0 to 1, u goes from -∞ to 0.

   I = ∫_(-∞)⁰ du/(u² + 2) = 1/√2[tan⁻¹ u/√2]_(-∞)⁰ = 1/√2(0 + π/2)

4. So. I = π/2√2, which is option (b).

The idea to keep. On symmetric limits, a factor like 1/(1+eᵒᵈᵈᶠᵘⁿᶜᵗⁱᵒⁿ) is a signal: swap x for -x and add. The two copies sum to 1, and you are left with half the plain integral.

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