Answer: (a) \(\frac13e^{3x+3}+C\); (b)(ii) \(\frac18-\frac{\pi\sqrt3}{6}\).

(a) Since

\((e^{x+1})^3=e^{3x+3},\)

we have

\(\int(e^{x+1})^3\,dx=\int e^{3x+3}\,dx.\)

Therefore

\(\int(e^{x+1})^3\,dx=\frac13e^{3x+3}+C.\)

(b)(i) Differentiate using the product rule:

\(\frac{d}{dx}(x\sin4x)=\sin4x+4x\cos4x.\)

(b)(ii) From part (b)(i),

\(4x\cos4x=\frac{d}{dx}(x\sin4x)-\sin4x.\)

So

\(\int4x\cos4x\,dx=x\sin4x+\frac14\cos4x.\)

Therefore

At \(x=\frac{\pi}{3}\),

At \(x=\frac{\pi}{4}\),

Subtracting gives