I saw this today, I checked in Mathematica and the integral comes out to $$\pi$$, but I have no idea how to solve it.
FREE Wi-Fi: The Wi-Fi password is the first $$10$$ digits of the answer. $$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$
The integrand is the sum of an odd and even function, and only the latter contributes, so it's $$\int_0^2\sqrt{4-x^2}dx$$. This is a quarter of the area of a radius-$$2$$ circle, i.e. $$\pi$$.
We have that $$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$ since the integrand is point symmetrix in the origin. Since $$\sqrt{4-x^2}$$ on $$[-2;2]$$ is the formula for the upper part of a circle we find that $$\int_{-2}^2 \sqrt{4-x^2} dx=\frac 12 \pi r^2=\frac 12 \pi 2^2=2\pi$$ So the whole integral is: $$\int_{-2}^2(x^3\cos \frac x2 +\frac 12 )\sqrt{4-x^2} dx=\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx+\frac 12\int_{-2}^2 \sqrt{4-x^2} dx=0+\frac 12 2\pi=\pi$$