Answer: (a) \(p=4,\ q=2,\ r=-4,\ s=-2\).

(b) \(\displaystyle \int_1^2 x^4\sqrt{4-x^2}\,dx=\frac{4\pi}{3}+\sqrt{3}\).

Start by expanding the algebraic expression in terms of powers of \(z\):

\(\left(z+\frac{1}{z}\right)^4= z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\)

and

\(\left(z-\frac{1}{z}\right)^2=z^2-2+\frac{1}{z^2}\).

It is quicker to regroup first:

\(\left(z+\frac{1}{z}\right)^4\left(z-\frac{1}{z}\right)^2=\left(z^2+2+\frac{1}{z^2}\right)\left(z^4-2+\frac{1}{z^4}\right).\)

Multiplying out gives

\(=z^6+2z^4-z^2-4-\frac{1}{z^2}+\frac{2}{z^4}+\frac{1}{z^6}.\)

Now set \(z=\cos\theta+i\sin\theta\), so \(z^n+z^{-n}=2\cos n\theta\). Hence

\(\left(z^6+\frac{1}{z^6}\right)+2\left(z^4+\frac{1}{z^4}\right)-\left(z^2+\frac{1}{z^2}\right)-4\)

becomes

\(2\cos6\theta+4\cos4\theta-2\cos2\theta-4.\)

On the other hand, the left-hand side is

\(\left(\cos\theta+i\sin\theta+\cos\theta-i\sin\theta\right)^4\left(\cos\theta+i\sin\theta-\cos\theta+i\sin\theta\right)^2\)

so equivalently, using \(z+z^{-1}=2\cos\theta\) and \(z-z^{-1}=2i\sin\theta\), we have

\(16\cos^4\theta\cdot(-4\sin^2\theta)=2\cos6\theta+4\cos4\theta-2\cos2\theta-4.\)

Therefore

\(64\sin^2\theta\cos^4\theta=4+2\cos2\theta-4\cos4\theta-2\cos6\theta,\)

so \(p=4, q=2, r=-4, s=-2\).

For the integral, use \(x=2\cos\theta\). Then \(dx=-2\sin\theta\,d\theta\), and when \(x=2\), \(\theta=0\); when \(x=1\), \(\theta=\pi/3\).

Also, \(x^4=(2\cos\theta)^4=16\cos^4\theta\) and

\(\sqrt{4-x^2}=\sqrt{4-4\cos^2\theta}=2\sin\theta\)

for \(0\le \theta\le \pi/3\). Hence

\(\displaystyle \int_1^2 x^4\sqrt{4-x^2}\,dx=\int_{\pi/3}^{0}16\cos^4\theta\,(2\sin\theta)(-2\sin\theta)\,d\theta\)

\(=\int_0^{\pi/3}64\cos^4\theta\sin^2\theta\,d\theta.\)

Using the trigonometric identity found above, this becomes

\(\displaystyle \int_0^{\pi/3}\left(4+2\cos2\theta-4\cos4\theta-2\cos6\theta\right)d\theta.\)

Integrating term by term:

\(\displaystyle \left[4\theta+\sin2\theta-\sin4\theta-\frac13\sin6\theta\right]_0^{\pi/3}.\)

Now \(\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\), \(\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}\), and \(\sin 2\pi=0\), so

\(\displaystyle \frac{4\pi}{3}+\frac{\sqrt3}{2}+\frac{\sqrt3}{2}=\frac{4\pi}{3}+\sqrt3.\)