Question

$\int_{-\pi}^\pi (1 - x^2)\sin x \cos^2 x \, dx =$

• A. $\frac{\pi}{3}$
• B. $2\pi - \pi^2$
• C. $\frac{\pi^3}{2}$
• D. $0$

Answer 1

The Correct Option is D

The function $(1 - x^2)\sin x \cos^2 x$ is odd because $(1 - x^2)$ is even and $\sin x$ is odd.

The integral of an odd function over $[-a, a]$ is zero: \[ \int_{-\pi}^\pi (1 - x^2)\sin x \cos^2 x \, dx = 0. \]