Answer: (a)(i) \(-3\cos\left(\frac{\phi+\pi}{3}\right)+C\); (a)(ii) \(5\theta+C\); (b) \(\frac{3e-1}{e}\).

(a)(i) Let

\(u=\frac{\phi+\pi}{3}.\)

Then \(du=\frac13\,d\phi\), so \(d\phi=3\,du\). Hence

Therefore

(a)(ii) Since \(\sin^2\theta+\cos^2\theta=1\),

\(5\sin^2\theta+5\cos^2\theta=5.\)

So

(b) First expand the integrand:

Therefore

Integrate:

\(\left[2\ln x-\frac1x\right]_1^e.\)

Substitute the limits:

Thus

\(3-\frac1e=\frac{3e-1}{e}.\)