Question

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

Answer 1

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

Applying integration by parts, taking

\[\sin\left( 3x + 1 \right)\] as first function and \[e^{2x}\]as second function, we get

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

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

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

Again applying integration by parts, taking

\[\cos\left( 3x + 1 \right)\] as first function and

\[e^{2x}\]as second function, we get

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

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

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

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

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

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

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