The integral provided in the image is:

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

To solve this integral, let's break it down into parts.

Symmetry Analysis

Notice that the integrand is composed of two main parts:

1. $x^3 \cos\left(\frac{x}{2}\right)$
2. $\frac{1}{2} \sqrt{4 - x^2}$

Part 1: $x^3 \cos\left(\frac{x}{2}\right) \sqrt{4 - x^2}$

This part is an odd function because it contains $x^3$ and $\cos\left(\frac{x}{2}\right)$, both of which preserve the odd nature when multiplied by the square root term (which is even).

An odd function integrated over a symmetric interval $[-a, a]$ results in zero:

$\int_{-2}^{2} x^3 \cos\left(\frac{x}{2}\right) \sqrt{4 - x^2} \, dx = 0$

Part 2: $\frac{1}{2} \sqrt{4 - x^2}$

This is an even function because it doesn't change sign if $x$ is replaced by $-x$.

So, the integral simplifies to:

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

Solving the Simplified Integral

The integral of $\sqrt{4 - x^2}$ is a well-known form and can be evaluated directly:

$\int \sqrt{a^2 - x^2} \, dx = \frac{a^2}{2} \left(\arcsin\left(\frac{x}{a}\right)\right) + \frac{x}{2} \sqrt{a^2 - x^2} + C$

For $a = 2$, the integral becomes:

$\int \sqrt{4 - x^2} \, dx = \frac{1}{2} \left(2^2 \arcsin\left(\frac{x}{2}\right) + x \sqrt{4 - x^2}\right) + C$

Substituting the limits from $-2$ to $2$:

$\int_{-2}^{2} \sqrt{4 - x^2} \, dx = \frac{1}{2} \left[2 \times \left(2 \times \arcsin\left(\frac{2}{2}\right) + 2 \sqrt{4 - 4}\right)\right] = \frac{1}{2} \times (2\pi \times 2) = 2\pi$

So, the final answer for the integral is:

$\text{Result} = \frac{1}{2} \times 2\pi = \pi$

Therefore, the integral evaluates to $\pi$.

Would you like further details or have any questions? Here are some related questions to consider:

1. What happens if the interval of integration is not symmetric around zero?
2. How would the solution change if the integrand included another odd function?
3. Can we compute this integral using polar coordinates?
4. What other functions are odd and vanish when integrated over a symmetric interval?
5. How does the substitution method work for integrals like this?

Tip: When solving definite integrals, always check for symmetry, as it can simplify the problem significantly.