If ∫_(-π/2)^(π/2) (96 x² cos ² x)/(1+eˣ) d x=π(α π²+β), α, β ∈ ℤ, then (α+β)² equals

Solution

Maths sir · 4 steps

Replace x by −x and add: the 1+eˣ in the denominator vanishes and the integral halves cleanly.

1. Use x to −x. Call the integral I. Put x → -x. The limits stay symmetric and 96x²cos² x is unchanged, so I = ∫_(-π/2)^(π/2) (96x²cos² x)/(1+e⁻ˣ) dx = ∫_(-π/2)^(π/2) (96x²cos² x eˣ)/(1+eˣ) dx.
2. Add the two forms. Adding the two forms of I gives (1+eˣ)/(1+eˣ) = 1, so 2I = ∫_(-π/2)^(π/2) 96x²cos² x dx. The integrand is even, so I = 96∫₀^(π/2) x²cos² x dx.

   ∫₀^(π/2) x²cos² x dx = 1/2∫₀^(π/2) x² dx + 1/2∫₀^(π/2) x²cos 2x dx = π³/48 + 1/2(-π/4)

3. By parts. Integrating by parts twice, ∫ x²cos 2x dx = (x²sin 2x)/2 + (x cos 2x)/2 - (sin 2x)/4. From 0 to π/2 this gives π/2·-1/2 = -π/4. So the integral is π³/48 - π/8.
4. So. I = 96(π³/48 - π/8) = 2π³ - 12π = π(2π² - 12). Then α = 2, β = -12, and (α+β)² = (-10)² = 100, option (b).

The idea to keep. When you see f(x)/(1+aˣ) on symmetric limits with f even, the answer is just ∫₀ᵃ f(x) dx. Remember this shortcut, it comes up again and again.

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