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AIEEE 2012

JEE Main 2012 Mathematics question on Definite Integrals and Areas

The value of the integral ∫−π/2π/2(x2+log⁡π−xπ+x)cos⁡x dx\int_{-\pi/2}^{\pi/2} \left(x^2 + \log\frac{\pi - x}{\pi + x}\right) \cos x \, dx is
  1. A.
    00
  2. B.
    π22−4\dfrac{\pi^2}{2} - 4
  3. C.
    π22+4\dfrac{\pi^2}{2} + 4
  4. D.
    π22\dfrac{\pi^2}{2}
Show answer and solution

Correct option: B

Solution

Split the integrand: x²cosx + log((π-x)/(π+x))·cosx.

The second term: log((π-x)/(π+x)) is an odd function (replacing x with -x gives log((π+x)/(π-x)) = -log((π-x)/(π+x))), and cosx is even, so their product is odd. Its integral over [-π/2, π/2] = 0.

The first term x²cosx is even, so the integral = 2∫₀^(π/2) x²cosx dx.

Integrate by parts twice: ∫x²cosx dx = x²sinx - 2∫x sinx dx = x²sinx - 2(-x cosx + sinx) + C = x²sinx + 2x cosx - 2sinx + C.

Evaluate from 0 to π/2: [(π/2)²·1 + 2·(π/2)·0 - 2·1] - [0] = π²/4 - 2.

Multiply by 2: π²/2 - 4.

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