The integral you provided is:

$\int_{-2}^{2} \left( x^3 \cos \left( \frac{x}{2} \right) + \frac{1}{2} \right) \sqrt{4 - x^2} \, dx$

Steps to solve:

1. Symmetry Consideration: The limits of integration are symmetric about zero (from $-2$ to $2$), and the integrand involves both even and odd functions:

   • $x^3 \cos\left( \frac{x}{2} \right)$ is an odd function (since $x^3$ is odd, and cosine is an even function).
   • $\frac{1}{2} \sqrt{4 - x^2}$ is an even function.

   Using this, we can simplify the integral:

   • The integral of an odd function over a symmetric interval $[-a, a]$ is zero.
   • So, the term involving $x^3 \cos \left( \frac{x}{2} \right)$ vanishes, and we are left with only the even part of the integrand:

   $\int_{-2}^{2} \frac{1}{2} \sqrt{4 - x^2} \, dx$

2. Evaluate the Remaining Integral: The integral we are now solving is:

   $\frac{1}{2} \int_{-2}^{2} \sqrt{4 - x^2} \, dx$

   This is a standard integral that represents the area of a semicircle with radius 2. The formula for the area of a semicircle of radius $R$ is:

   $\text{Area} = \frac{1}{2} \pi R^2$

   In this case, $R = 2$, so the area is:

   $\text{Area} = \frac{1}{2} \pi (2^2) = 2\pi$

3. Final Result: Therefore, the value of the original integral is:

   $\boxed{2\pi}$

Would you like more details on any part of this solution?

Related Questions:

1. How do you determine whether a function is odd or even?
2. What are some techniques to solve definite integrals using symmetry?
3. How do you compute the integral of $\sqrt{4 - x^2}$ geometrically?
4. Can this integral be solved using a trigonometric substitution instead of geometry?
5. What is the role of the cosine function in integrals involving oscillatory behavior?

Tip:

When solving integrals with symmetric limits, always check for odd and even functions. This can significantly simplify your work!