The value of the integral
∫_(- π /2)^(π /2) dx/(1 + eˣ)(sin ⁶x + cos ⁶x) is equal to

Solution

Maths sir · 4 steps

Pair x with −x to kill the eˣ, then divide by cos⁴ x and turn it into a tan integral.

1. Remove the e^x. Let f(x) = 1/(sin⁶ x + cos⁶ x), which is even. Replacing x by -x turns 1/(1+eˣ) into eˣ/(1+eˣ). Adding the two forms of I makes this factor add up to 1, so 2I = ∫_(-π/2)^(π/2) f(x) dx = 2∫₀^(π/2) f(x) dx.
2. Convert to tan. Use sin⁶ x + cos⁶ x = 1 - 3 sin² x cos² x. Divide the top and the bottom by cos⁴ x and put t = tan x, so sec² x dx = dt. The bottom becomes (1+t²)² - 3t² = t⁴ - t² + 1, and the top becomes (1+t²) dt.

   I = ∫₀^∞ (1+t²)/(t⁴ - t² + 1) dt = ∫₀^∞ (1 + 1/t²)/((t - 1/t)² + 1) dt

3. Substitute again. Put u = t - 1/t, so du = (1 + 1/t²)dt. As t goes from 0 to ∞, u goes from -∞ to ∞. This gives I = ∫_(-∞)^∞ du/(u²+1) = [tan⁻¹ u]_(-∞)^∞ = π.
4. So. The integral equals π, which is option (c).

The idea to keep. When 1/(1+eˣ) multiplies an even function over [-a, a], it drops out: the integral equals ∫₀ᵃ of the even function. Keep this shortcut ready.

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