Question:

<p><strong>Step 1:</strong> Decompose the integrand into even and odd parts:</p><p>Let f(x) = x + x². We write:</p><ul><li>x·cos(nx) is an <strong>odd</strong> function (odd × even = odd)</li><li>x²·cos(nx) is an <strong>even</strong> function (even × even = even)</li></ul><p><strong>Step 2:</strong> Apply the property that integral of odd functions over symmetric intervals is zero:</p><p>$$a_n = \frac{1}{\pi}\left[\int_{-\pi}^{\pi} x\cos(nx)\,dx + \int_{-\pi}^{\pi} x^2\cos(nx)\,dx\right]$$</p><p>$$a_n = \frac{1}{\pi}\left[0 + \int_{-\pi}^{\pi} x^2\cos(nx)\,dx\right]$$</p><p><strong>Step 3:</strong> Evaluate the even part using integration by parts twice (or use symmetry to write as):</p><p>$$a_n = \frac{2}{\pi}\int_{0}^{\pi} x^2\cos(nx)\,dx$$</p><p><strong>Step 4:</strong> Apply integration by parts:</p><p>Let u = x², dv = cos(nx)dx → du = 2x dx, v = sin(nx)/n</p><p>$$\int_0^\pi x^2\cos(nx)\,dx = \left[\frac{x^2\sin(nx)}{n}\right]_0^\pi - \int_0^\pi \frac{2x\sin(nx)}{n}\,dx$$</p><p>The boundary term vanishes (sin(nπ) = 0). Repeat integration by parts on the remaining integral to get:</p><p>$$= \frac{4}{n^2}(-1)^n$$</p><p><strong>Step 5:</strong> Substitute back:</p><p>$$a_n = \frac{2}{\pi} \cdot \frac{4(-1)^n}{n^2} = \frac{8(-1)^n}{\pi n^2}$$</p><p>∴ Answer: <strong>A</strong> (assuming option A is $\frac{8(-1)^n}{\pi n^2}$ for n ≠ 0, and a₀ = π²/3)</p>