Answer: \(\sin^6\theta=\frac{1}{32}(10-15\cos 2\theta+6\cos 4\theta-\cos 6\theta)\).

Hence \(p=10\), \(q=-15\), \(r=6\) and \(s=-1\).

Also, \(\displaystyle \int_0^{\pi/4}\sin^6\theta\,d\theta=\frac{5\pi}{64}-\frac{11}{48}\).

Let \(z=\cos\theta+i\sin\theta\). Then \(z^{-1}=\cos\theta-i\sin\theta\), so

\(z-z^{-1}=2i\sin\theta\), and hence \((z-z^{-1})^6=(2i\sin\theta)^6=-64\sin^6\theta\).

Now expand:

\(\begin{aligned}(z-z^{-1})^6&=z^6-6z^4+15z^2-20+15z^{-2}-6z^{-4}+z^{-6}\\&=(z^6+z^{-6})-6(z^4+z^{-4})+15(z^2+z^{-2})-20.\end{aligned}\)

Using \(z^n+z^{-n}=2\cos n\theta\),

\(\begin{aligned}(z-z^{-1})^6&=2\cos6\theta-12\cos4\theta+30\cos2\theta-20.\end{aligned}\)

So

\(\begin{aligned}-64\sin^6\theta&=2\cos6\theta-12\cos4\theta+30\cos2\theta-20,\\64\sin^6\theta&=20-30\cos2\theta+12\cos4\theta-2\cos6\theta,\\\sin^6\theta&=\frac{1}{32}(10-15\cos2\theta+6\cos4\theta-\cos6\theta).\end{aligned}\)

Thus \(p=10\), \(q=-15\), \(r=6\), \(s=-1\).

Now integrate:

\(\begin{aligned}\int_0^{\pi/4}\sin^6\theta\,d\theta&=\frac{1}{32}\int_0^{\pi/4}(10-15\cos2\theta+6\cos4\theta-\cos6\theta)\,d\theta\\&=\left[\frac{5\theta}{16}-\frac{15\sin2\theta}{64}+\frac{3\sin4\theta}{64}-\frac{\sin6\theta}{192}\right]_0^{\pi/4}.\end{aligned}\)

At \(\theta=\frac{\pi}{4}\), \(\sin\frac{\pi}{2}=1\), \(\sin\pi=0\), \(\sin\frac{3\pi}{2}=-1\). At \(\theta=0\), all sine terms are \(0\). Therefore

\(\begin{aligned}\int_0^{\pi/4}\sin^6\theta\,d\theta&=\frac{5\pi}{64}-\frac{15}{64}+0-\frac{-1}{192}\\&=\frac{5\pi}{64}-\frac{15}{64}+\frac{1}{192}\\&=\frac{5\pi}{64}-\frac{11}{48}.\end{aligned}\)