The value of the integral given below is int_{-pi}^{pi} x^2 cos(x) dx A. -2pi B. pi C. -pi D. 2pi

GATE 2014 · Engineering Mathematics · Integration · medium

Answer: -2pi

1. Use even symmetry: I = 2 * int_0^pi x^2 cos(x) dx
2. Integration by parts twice: int x^2 cos(x) dx: u=x^2, dv=cos(x)dx => x^2 sin(x) - int 2x sin(x) dx. Then: u=2x, dv=sin(x)dx => x^2 sin(x) - [-2x cos(x) + int 2 cos(x) dx] = x^2 sin(x) + 2x cos(x) - 2 sin(x).
3. Evaluate at bounds and multiply: F(pi) = pi^2 sin(pi) + 2*pi*cos(pi) - 2 sin(pi) = 0 + 2*pi*(-1) - 0 = -2*pi. F(0) = 0. I = 2*(-2*pi - 0) = -4*pi... wait: I = 2*[F(pi)-F(0)] = 2*(-2pi) = -4pi.