Don't go by looks

This is an integral of a function over an interval symmetric about the origin. That makes the problem easy to deal with, even though finding a formula for an anti-derivative might be quite a task.

First, consider the various parts of the function.

• $x^3$ is an odd function

• $\cos (\frac{x}{2})$ is an even function

• $\sqrt{4-x^2}$ is an even function.

• $\frac{1}{2}$ is an even function.

To simplify, break the integral into two parts:

$\int_{-2}^{2} x^3 \cos(\frac{x}{2}) \sqrt{4-x^2} dx,$ and

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

The integrand in the first integral is the product of two even functions and an odd function. Thus it is an odd function. And thus its integral over [-2,2] equals zero.

Thus the value of the integral as a whole reduces to the value of the integral of its even part, which itself equals 1/2 the area of the upper semi-circle of radius 2 centered at the origin. Thus

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

$\begin{aligned} I & = \int_{-2}^2 \left(x^3\cos \frac x2 + \frac 12\right) \sqrt{4-x^2}\ dx \\ & = {\color{#3D99F6}\int_{-2}^2 x^3\cos \frac x2 \sqrt{4-x^2}\ dx} + \color{#D61F06} \int_{-2}^2 \frac 12 \sqrt{4-x^2}\ dx & \small \color{#3D99F6} \text{Since the first integrand is odd} \\ & = {\color{#3D99F6}0} + \color{#D61F06} \int_0^2 \sqrt{4-x^2}\ dx & \small \color{#D61F06} \text{and the second integrand is even} \\ & = \int_0^2 2 \sqrt {1-\frac {x^2}4}\ dx & \small \color{#3D99F6} \text{Let }\frac x2 = \sin \theta \implies \frac 12 dx = \cos \theta \ d\theta \\ & = 4 \int_0^\frac \pi 2 \cos^2 \theta \ d\theta \\ & = 2 \int_0^\frac \pi 2 (1+\cos 2\theta) \ d\theta \\ & = 2 \left[ \theta + \frac {\sin 2\theta}2 \right]_0^\frac \pi 2 \\ & = \boxed \pi \end{aligned}$