Answer: (i) \(16\sin^5\theta=\sin 5\theta-5\sin 3\theta+10\sin\theta\), so \(p=-5\) and \(q=10\).

(ii) \(\displaystyle \int_0^{\pi/3} 16\sin^5\theta\,d\theta=\frac{53}{30}\).

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

\(z-\frac{1}{z}=(\cos\theta+\mathrm{i}\sin\theta)-(\cos\theta-\mathrm{i}\sin\theta)=2\mathrm{i}\sin\theta.\)

Now raise both sides to the fifth power:

\(\left(z-\frac{1}{z}\right)^5=(2\mathrm{i}\sin\theta)^5=32\mathrm{i}\sin^5\theta.\)

Also, expanding the left-hand side gives

\(\left(z-\frac{1}{z}\right)^5=z^5-5z^3\frac{1}{z}+10z\frac{1}{z^2}-10\frac{1}{z}\frac{1}{z^2}+5\frac{1}{z^3}-\frac{1}{z^5},\)

which can be regrouped as

\(\left(z-\frac{1}{z}\right)^5=\left(z^5-\frac{1}{z^5}\right)-5\left(z^3-\frac{1}{z^3}\right)+10\left(z-\frac{1}{z}\right).\)

Since \(z^n- z^{-n}=2\mathrm{i}\sin(n\theta)\), this becomes

\(32\mathrm{i}\sin^5\theta=2\mathrm{i}\sin5\theta-10\mathrm{i}\sin3\theta+20\mathrm{i}\sin\theta.\)

Dividing by \(2\mathrm{i}\) gives

\(16\sin^5\theta=\sin5\theta-5\sin3\theta+10\sin\theta.\)

Hence \(p=-5\) and \(q=10\).

(ii) Integrate term by term:

\(\displaystyle \int_0^{\pi/3} 16\sin^5\theta\,d\theta=\int_0^{\pi/3}(\sin5\theta-5\sin3\theta+10\sin\theta)\,d\theta.\)

So

\(\displaystyle =\left[-\frac{\cos5\theta}{5}+\frac{5\cos3\theta}{3}-10\cos\theta\right]_0^{\pi/3}.\)

At \(\theta=\pi/3\), \(\cos(5\pi/3)=\tfrac12\), \(\cos\pi=-1\), and \(\cos(\pi/3)=\tfrac12\), so the value is

\(\displaystyle -\frac{1}{10}-\frac{5}{3}-5.\)

At \(\theta=0\), the value is

\(\displaystyle -\frac{1}{5}+\frac{5}{3}-10.\)

Therefore

\(\displaystyle \left(-\frac{1}{10}-\frac{5}{3}-5\right)-\left(-\frac{1}{5}+\frac{5}{3}-10\right)=\frac{53}{30}.\)

So the exact value of the integral is \(\frac{53}{30}\).