Question

Evaluate the following integrals:

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

Answer 1

\[\text{ We have, } \]
\[I = \int e^{2x} \sin\left( 3x + 1 \right) dx\]
\[\text{Let the first function be sin ( 3x + 1 ) and the second function be} \text{ e}^{2x} . \]
\[\text{First we find the integral of the second function}, i . e . , \int e^{2x} \text{ dx }. \]
\[\int e^{2x} dx = \frac{1}{2} e^{2x} \]
\[\text{Now, using integration by parts, we get}\]
\[I = \text{ sin}\left( 3x + 1 \right)\int e^{2x} dx - \int\left[ \left( \frac{d \left( \sin\left( 3x + 1 \right) \right)}{d x} \right)\int e^{2x} dx \right]dx\]
\[ = \frac{1}{2} \text{ sin}\left( 3x + 1 \right) e^{2x} - \frac{3}{2}\int\left[ \cos\left( 3x + 1 \right) e^{2x} \right]dx\]
\[ = \frac{1}{2} \text{ sin}\left( 3x + 1 \right) e^{2x} - \frac{3}{2}\left\{ \cos\left( 3x + 1 \right)\int e^{2x} \text{ dx }- \int\left[ \left( \frac{d \left( \cos\left( 3x + 1 \right) \right)}{d x} \right)\int e^{2x} dx \right]\text{ dx }\right\}\]
\[ = \frac{1}{2}\text{ sin }\left( 3x + 1 \right) e^{2x} - \frac{3}{2}\left\{ \frac{1}{2}\cos\left( 3x + 1 \right) e^{2x} + \frac{3}{2}\int\text{ sin}\left( 3x + 1 \right) e^{2x} dx \right\}\]
\[ = \frac{1}{2}\text{ sin }\left( 3x + 1 \right) e^{2x} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} - \frac{9}{4}I + c\]
\[I + \frac{9}{4}I = \frac{1}{2}\text{ sin }\left( 3x + 1 \right) e^{2x} - \frac{3}{4}\cos\left( 3x + 1 \right) e^{2x} + c\]
\[\frac{13}{4}I = \frac{e^{2x}}{2}\left[ \text{ sin }\left( 3x + 1 \right) - \frac{3}{2}\cos\left( 3x + 1 \right) \right] + c\]
\[I = \frac{2}{13} e^{2x} \left[ \text{ sin }\left( 3x + 1 \right) - \frac{3}{2}\cos\left( 3x + 1 \right) \right] + c\]
\[ = \frac{e^{2x}}{13}\left[ 2 \text{ sin }\left( 3x + 1 \right) - 3 \cos\left( 3x + 1 \right) \right] + c\]
\[\text{ Hence, } \int e^{2x} \text{ sin }\left( 3x + 1 \right) dx = \frac{e^{2x}}{13}\left[ 2 \text{ sin }\left( 3x + 1 \right) - 3 \cos\left( 3x + 1 \right) \right] + c\]