Evaluate the definite integral from x = -2 to x = 2 of (x^3 cos(x/2) + 1/2) times the square root of (4 - x^2).

Step-by-step solution

1. 1

   Split the integral before trying to antidifferentiate. The factor $x^{3}\cos\frac{x}{2}\sqrt{4-x^{2}}$ has no elementary antiderivative, so any attempt at substitution or parts is doomed. Because the interval is symmetric about $0$, split instead:

   $\int_{-2}^{2}x^{3}\cos\frac{x}{2}\sqrt{4-x^{2}}\,dx+\frac{1}{2}\int_{-2}^{2}\sqrt{4-x^{2}}\,dx.$

2. 2

   Classify the first piece by parity. $x^{3}$ is odd; $\cos\frac{x}{2}$ is even; $\sqrt{4-x^{2}}$ is even. Odd $\times$ even $\times$ even is odd, so the whole integrand satisfies $f(-x)=-f(x)$. On a symmetric interval every odd function integrates to zero, because the signed area on $[-2,0]$ exactly cancels the signed area on $[0,2]$:

   $\int_{-2}^{2}x^{3}\cos\frac{x}{2}\sqrt{4-x^{2}}\,dx=0.$

   This is the whole point of the problem — the hard-looking term is a decoy.

3. 3

   Recognise the second piece as a semicircle. The curve $y=\sqrt{4-x^{2}}$ is equivalent to $x^{2}+y^{2}=4$ with $y\ge 0$: the upper half of a circle of radius $2$. So the integral is a geometric area, not something to antidifferentiate:

   $\int_{-2}^{2}\sqrt{4-x^{2}}\,dx=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi(2)^{2}=2\pi.$

4. 4

   Reassemble. Multiply the semicircle by the constant $\tfrac12$ and add the vanished odd part:

   $\int_{-2}^{2}\left(x^{3}\cos\frac{x}{2}+\frac{1}{2}\right)\sqrt{4-x^{2}}\,dx=0+\frac{1}{2}(2\pi)=\pi.$

5. 5

   Check numerically. Composite Simpson's rule on $[-2,2]$ gives $3.1415926$ for the full integrand and $2\times10^{-16}$ for the odd part alone, confirming both the exact value $\pi\approx 3.14159$ and the cancellation argument.