Final answer:
The definite integral involves an odd function over a symmetric interval, thus the integral evaluates to 0 due to the properties of symmetry.
Step-by-step explanation:
The student has presented a definite integral to be evaluated, which involves trigonometric and algebraic functions within a square root. To approach this integral, we observe the symmetry of the integrand and the limits of integration. Specifically, since the integral is evaluated from −2 to 2, we can consider whether the function is odd or even.
This integrand is structured as a product of two parts: the first is a quadratic function times the cosine function, and the second part involves a square root of a quadratic expression. By examining each part, it is evident that the cosine function is an even function while the x³ term is an odd function. Multiplying an even by an odd function results in an odd function. Consequently, when the limits of integration are symmetric about the y-axis, as they are in this case, the integral of an odd function over this interval is zero.
Furthermore, the square root part √(4-x²) maintains symmetry about the y-axis. As such, it does not affect the odd nature of the other part of the function. Therefore, the integral over the symmetric interval results in a value of 0, as the areas above and below the x-axis cancel each other out.