Problem Statement
Evaluate:\n\n$$\int e^{2x} \sin(3x) \, dx$$
Verified Solution & Marking Scheme
Set up Integration by Parts
Let $$I = \int e^{2x} \sin(3x) \, dx$$. Using $$\int u v' \, dx = u v - \int u' v \, dx$$, let $$u = \sin(3x)$$ and $$v' = e^{2x}$$.\nThen $$u' = 3\cos(3x)$$ and $$v = \frac{1}{2}e^{2x}$$.\n\n$$I = \frac{1}{2} e^{2x} \sin(3x) - \frac{3}{2} \int e^{2x} \cos(3x) \, dx$$
Apply by parts to the second integral
For $$\int e^{2x} \cos(3x) \, dx$$, let $$u_2 = \cos(3x)$$ and $$v_2' = e^{2x}$$.\nThen $$u_2' = -3\sin(3x)$$ and $$v_2 = \frac{1}{2}e^{2x}$$.\n\n$$\int e^{2x} \cos(3x) \, dx = \frac{1}{2} e^{2x} \cos(3x) + \frac{3}{2} \int e^{2x} \sin(3x) \, dx = \frac{1}{2} e^{2x} \cos(3x) + \frac{3}{2} I$$
Substitute back into equation for I
$$I = \frac{1}{2} e^{2x} \sin(3x) - \frac{3}{2} \left( \frac{1}{2} e^{2x} \cos(3x) + \frac{3}{2} I \right)$$\n$$I = \frac{1}{2} e^{2x} \sin(3x) - \frac{3}{4} e^{2x} \cos(3x) - \frac{9}{4} I$$\n$$I + \frac{9}{4} I = \frac{13}{4} I = \frac{e^{2x}}{4} (2\sin(3x) - 3\cos(3x))$$
Solve for I
$$I = \frac{e^{2x}}{13} (2\sin(3x) - 3\cos(3x)) + C$$