Answer: (i) \(I_n+I_{n-1}=\frac2{2n-1}\).

(ii) \(\int_0^{\pi/2}\frac{\sin8\theta}{\cos\theta}\,d\theta=-\frac{152}{105}\).

Let \(I_n=\int_0^{\pi/2} \frac{\sin 2n\theta}{\cos\theta}\,d\theta\).

(i) For positive integers \(n\),

\(I_n+I_{n-1}=\int_0^{\pi/2} \frac{\sin 2n\theta+\sin(2n-2)\theta}{\cos\theta}\,d\theta.\)

Using \(\sin P+\sin Q=2\sin\frac{P+Q}{2}\cos\frac{P-Q}{2}\) with \(P=2n\theta\) and \(Q=(2n-2)\theta\),

\(\sin 2n\theta+\sin(2n-2)\theta=2\sin(2n-1)\theta\cos\theta.\)

So

\(I_n+I_{n-1}=\int_0^{\pi/2} \frac{2\sin(2n-1)\theta\cos\theta}{\cos\theta}\,d\theta=\int_0^{\pi/2} 2\sin(2n-1)\theta\,d\theta.\)

Hence

\(I_n+I_{n-1}= \left[-\frac{2\cos(2n-1)\theta}{2n-1}\right]_0^{\pi/2}.\)

Now \(\cos\big((2n-1)\pi/2\big)=0\) and \(\cos 0=1\), so

\(I_n+I_{n-1}=0-\left(-\frac{2}{2n-1}\right)=\frac{2}{2n-1}.\)

(ii) We need \(I_4\), since \(\sin 8\theta=\sin(2\cdot 4\theta)\).

First find \(I_1\):

\(I_1=\int_0^{\pi/2} \frac{\sin 2\theta}{\cos\theta}\,d\theta=\int_0^{\pi/2} 2\sin\theta\,d\theta=[-2\cos\theta]_0^{\pi/2}=2.\)

Now use the recurrence from part (i):

\(I_2=\frac{2}{3}-I_1=\frac{2}{3}-2=-\frac{4}{3},\)

\(I_3=\frac{2}{5}-I_2=\frac{2}{5}+\frac{4}{3}=\frac{26}{15},\)

\(I_4=\frac{2}{7}-I_3=\frac{2}{7}-\frac{26}{15}=-\frac{152}{105}.\)

Therefore

\(\int_0^{\pi/2} \frac{\sin 8\theta}{\cos\theta}\,d\theta=-\frac{152}{105}.\)