Question

Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .

Answer 1

Let I =  \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] Use integration by parts,

\[\int u v dx = u\int v dx - \int\left[ \frac{du}{dx}\int v dx \right]dx\]

Here,

\[u = \sin \left( 3x + 1 \right) and v = e^{2x}\]

Therefore,

\[I = \sin \left( 3x + 1 \right)\int e^{2x} dx - \int\left[ \frac{d\left( \sin\left( 3x + 1 \right) \right)}{dx}\int e^{2x} dx \right]dx\]

\[I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{2}\int e^{2x} \cos\left( 3x + 1 \right) dx\]

\[I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{2}\left[ \cos\left( 3x + 1 \right)\int e^{2x} dx - \int\left\{ \frac{d\left( \cos\left( 3x + 1 \right) \right)}{dx}\int e^{2x} dx \right\} \right] \left[ \text {  Integration by parts again } \right]\]

\[I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{2}\left[ \frac{\cos\left( 3x + 1 \right) e^{2x}}{2} - \int\left\{ \frac{- 3}{2} e^{2x} \sin\left( 3x + 1 \right)dx \right\} \right]\]

\[I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} - \frac{9}{4}\int e^{2x} \sin\left( 3x + 1 \right)dx\]

\[I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} - \frac{9}{4}I\]

\[I + \frac{9}{4}I = \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} \]

\[I = \frac{4}{13}\left[ \frac{\sin \left( 3x + 1 \right) e^{2x}}{2} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} \right] + C\]