This is somewhere between high school to intro-level college calculus. The thing to notice is that the integral of odd-functions from -a to +a (a is any real number) is always zero. This is because the odd-function would have some value from -a to 0 and its negative from 0 to a, making the integral 0.
So only the integral of the second term is needed, which is easy to integrate after substituting x = 2 sin(\theta).
The answer is pi. The integral of first term is x^3 cos (x/2) * \sqrt(4 - x^2) is going to be 0 because sqrt(4 -x^2) is always positive, cos(x/2) in that range is also positive and x^3 will go from negative to positive making the area 0.
So, we just need [1/2 * \sqrt(4 - x^2)] from -2 to 2. Use x = 2sin\theta, and you get [2 cos^2(\theta)] from -pi/2 to pi/2 which turns out to be pi.
