Evaluate the definite integral from x = -2 to x = 2 of (x² cos(x/2) + 1/2) times √(4 - x²), and give a numerical value.

Step-by-step solution

1. 1

   Split the integral at the plus sign. The two pieces behave completely differently:

   $I=\underbrace{\int_{-2}^{2}x^{2}\cos\frac{x}{2}\sqrt{4-x^{2}}\,dx}_{J}+\frac12\int_{-2}^{2}\sqrt{4-x^{2}}\,dx.$

   Separating them first prevents mixing an elementary area with a genuinely special-function integral.

2. 2

   Evaluate the easy half geometrically. The graph of $y=\sqrt{4-x^{2}}$ on $[-2,2]$ is the upper half of the circle of radius $2$, so

   $\int_{-2}^{2}\sqrt{4-x^{2}}\,dx=\frac{\pi(2)^{2}}{2}=2\pi,\qquad \frac12(2\pi)=\pi.$

3. 3

   Recognise $J$ as a Bessel moment. Poisson's integral gives, for $z=ab$,

   $\int_{-a}^{a}\sqrt{a^{2}-x^{2}}\cos(bx)\,dx=\frac{\pi a}{b}J_{1}(ab).$

   Differentiating twice in $b$ brings down $-x^{2}$, so with $a=2$, $b=\tfrac12$, $z=1$ and the identities $\frac{d}{dz}\bigl(z^{-1}J_{1}\bigr)=-z^{-1}J_{2}$ and $\frac{d}{dz}\bigl(z^{-2}J_{2}\bigr)=-z^{-2}J_{3}$,

   $J=\pi a^{4}\left(\frac{J_{2}(z)}{z^{2}}-\frac{J_{3}(z)}{z}\right)=16\pi\bigl(J_{2}(1)-J_{3}(1)\bigr).$

4. 4

   Simplify with the recurrence. Since $J_{3}(1)=4J_{2}(1)-J_{1}(1)$, the same quantity can be written

   $J=16\pi\bigl(J_{1}(1)-3J_{2}(1)\bigr).$

   With $J_{1}(1)=0.4400506$ and $J_{2}(1)=0.1149035$ this gives $J=16\pi(0.0953401)=4.7923177$.

5. 5

   Add the two halves.

   $I=\pi+16\pi\bigl(J_{2}(1)-J_{3}(1)\bigr)=3.1415927+4.7923177=7.9339103.$

   High-precision numerical quadrature of the original integrand returns $7.933910333308135$, matching to fifteen digits. A frequent wrong route is to quote $J=8\pi J_{2}(1)-4\pi J_{1}(1)$, which evaluates to $-2.6420$ and would give $I\approx 0.4976$ — the wrong sign and the wrong magnitude, because that combination is not the second $b$-derivative of Poisson's formula.