Problem Statement
Consider the integral $I = \int e^{2x} \sin(3x) \, dx$.
**(a)** [5 marks] Using integration by parts twice, find an expression for $I$ in terms of $x$ and an arbitrary constant $C$.
**(b)** [2 marks] Hence, evaluate $\int_0^{\pi/3} e^{2x} \sin(3x) \, dx$, giving your answer in the form $\frac{a + b e^{2\pi/3}}{c}$ where $a, b, c \in \mathbb{Z}$.
Verified Solution & Marking Scheme
Part (a) [M1]
Set $u = \sin(3x) \implies du = 3\cos(3x)dx$ and $dv = e^{2x}dx \implies v = \frac{1}{2}e^{2x}$
Part (a) [A1]
$I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}\int e^{2x}\cos(3x)\,dx$
Part (a) [M1]
Apply parts again to $\int e^{2x}\cos(3x)dx$ with $u_2 = \cos(3x), dv_2 = e^{2x}dx \implies v_2 = \frac{1}{2}e^{2x}, du_2 = -3\sin(3x)dx$
Part (a) [A1]
$I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2}\left[\frac{1}{2}e^{2x}\cos(3x) + \frac{3}{2}\int e^{2x}\sin(3x)dx\right] = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{4}e^{2x}\cos(3x) - \frac{9}{4}I$
Part (a) [A1]
Rearranging: $\frac{13}{4}I = \frac{1}{4}e^{2x}[2\sin(3x) - 3\cos(3x)] \implies I = \frac{e^{2x}}{13}(2\sin(3x) - 3\cos(3x)) + C$
Part (b) [M1]
Substitute limits $0$ and $\frac{\pi}{3}$: at $x = \pi/3$, $\sin(\pi) = 0, \cos(\pi) = -1 \implies \frac{e^{2\pi/3}}{13}(0 - 3(-1)) = \frac{3e^{2\pi/3}}{13}$
Part (b) [A1]
At $x = 0$: $\frac{e^0}{13}(0 - 3(1)) = -\frac{3}{13}$. Subtracting yields $\frac{3 + 3e^{2\pi/3}}{13}$