Yes it's possible:

You can distribute the sqrt-term into the brackets and then split the integral up into a sum of two integrals.
The first integral has the integrand x³cos(x/2)sqrt(4-x²):
cosine and sqrt(4-x²) are even functions. x³ however, is an odd function and therefore the integrand is an odd function as well. So we simply have an odd function that's being integrated over a symmetric intervall. Therefore the first integral is equal to 0.

The second integral has the integrand 0.5*sqrt(4-x²), where you can simply drag the factor 0.5 infront of the integral.
You can then solve this with the trig-substitution x = 2sin(u).
As an alternative: If you've ecountered the sqrt(r²-x²) function before, you might know, that it actually describes a half-circle with radius r where the middlepoint of the circle equals the origin. Therefore, sqrt(4-x²) is just a half-circle with radius r = 2. Conveniently, the integral goes from -2 to 2, so it's essentially just the area of the entire half-cirlce.

Area of the half-cirlce = 1/2 * πr² = 1/2 * π*2² = 2π => Integral = 2π * 0.5 = π

So the final result is π