Solve this integral for free WiFi I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.

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  $$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$



We have that
$$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$
since the integrand is point symmetric 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:
$$\int_{-2}^2(x^3\cos \frac x2 +\frac 12 )\sqrt{4-x^2} dx=\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx+\frac 12\int_{-2}^2 \sqrt{4-x^2} dx=0+\frac 12 2\pi=\pi$$
