Even And Odd Functions

Symmetry of the integration interval [-\pi, \pi] invites splitting the integrand by parity before any computation, because the parity of each term determines its contribution. The governing identities are that for any odd function g, \int_{-a}^{a} g(x),dx = 0, while for any even function h, \int_{-a}^{a} h(x),dx = 2\int_0^a h(x),dx. These hold because the graph of an odd function cancels across the origin while an even function mirrors itself. Here x^3 is odd, contributing exactly 0 to the integral. The term \cos x is even, contributing 2\int_0^\pi \cos x,dx = 2[\sin x]_0^\pi = 2(\sin\pi - \sin 0) = 2(0 - 0) = 0. Therefore the total integral is 0 + 0 = 0. Option 2\pi would arise only if cosine were replaced by a nonzero constant integrated over the full width. Option \pi mistakes the cosine contribution for the interval length. Option 2 incorrectly evaluates \int_0^\pi \cos x as 1 rather than 0. As a final plausibility check, \sin\pi = 0 independently forces the even part to vanish, so both the parity argument and direct evaluation agree that the integral is genuinely zero across the symmetric domain.