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JEE Advanced 2016

JEE Main 2016 Mathematics question on Definite Integrals and Areas

The value of ∫−π/2π/2x2cos⁡x1+exdx\int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1 + e^x} dx is equal to
  1. A.
    π24−2\frac{\pi^2}{4} - 2
  2. B.
    π24+2\frac{\pi^2}{4} + 2
  3. C.
    π24−e−π/2\frac{\pi^2}{4} - e^{-\pi/2}
  4. D.
    π24+eπ/2\frac{\pi^2}{4} + e^{\pi/2}
Show answer and solution

Correct option: A

Solution

Use the property: for f(x) = x²cosx/(1+eˣ), write f(x) + f(-x) = x²cosx/(1+eˣ) + x²cosx/(1+e⁻ˣ) = x²cosx · [1/(1+eˣ) + eˣ/(eˣ+1)] = x²cosx.

So the integral I = (1/2)∫₋π/2^π/2 x²cosx dx = ∫₀^π/2 x²cosx dx.

Integrate by parts twice: ∫₀^π/2 x²cosx dx = [x²sinx]₀^π/2 - 2∫₀^π/2 x sinx dx = π²/4 - 2([−x cosx]₀^π/2 + ∫₀^π/2 cosx dx) = π²/4 - 2(0 + [sinx]₀^π/2) = π²/4 - 2.

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