Question

\[\int e^{2x} \cos \left( 3x + 4 \right) \text{ dx }\]

Answer 1

\[\text{ Let I } = \int e^{2x} \cos \left( 3x + 4 \right)dx\]
`\text{Considering cos  (3x + 4  ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[I = \text{ cos }\left( 3x + 4 \right)\frac{e^{2x}}{2} - \int - \text{ sin }\left( 3x + 4 \right) \times 3\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I = \frac{e^{2x} \text{ cos }\left( 3x + 4 \right)}{2} + \frac{3}{2}\int e^{2x} \text{ sin }\left( 3x + 4 \right)dx\]
\[ \Rightarrow I = \frac{e^{2x} \text{ cos } \left( 3x + 4 \right)}{2} + \frac{3}{2} I_1 . . . . . \left( 1 \right)\]
\[\text{ where I}_1 = \int e^{2x} \text{ sin } \left( 3x + 4 \right)dx\]
`\text{Considering cos  (3x + 4  ) as first function and` `\text{ e}^{2x}`   ` \text{ as second function} `
\[ I_1 = \text{ sin } \left( 3x + 4 \right)\frac{e^{2x}}{2} - \int 3 \text{ cos }\left( 3x + 4 \right)\frac{e^{2x}}{2}dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \text{ sin } \left( 3x + 4 \right)}{2} - \frac{3}{2}\int e^{2x} \text{ cos} \left( 3x + 4 \right)dx\]
\[ \Rightarrow I_1 = \frac{e^{2x} \text{ sin }\left( 3x + 4 \right)}{2} - \frac{3}{2} I . . . . . \left( 2 \right)\]
` \text{ From ( 1 ) and ( 2 ) } `
\[I = \frac{e^{2x} \text{ cos }\left( 3x + 4 \right)}{2} + \frac{3}{4} \text{ e}^{2x} \text{ sin }\left( 3x + 4 \right) - \frac{9}{4}I\]
\[ \Rightarrow I + \frac{9}{4}I = \frac{2 e^{2x} \cos\left( 3x + 4 \right) + 3 e^{2x} \text{ sin }\left( 3x + 4 \right)}{4}\]
\[ \Rightarrow I = \frac{e^{2x}}{13}\left[ 2 \text{ cos } \left( 3x + 4 \right) + 3 \text{ sin }\left( 3x + 4 \right) \right] + C\]