Question

\[\int e^{ax} \cos\ bx\ dx\]

Answer 1

\[\text{ Let I } = \int e^{ax} \cos\left( bx \right)\text{ dx }\]
`  \text{Considering cos (  bx ) as first function and }`  `  e^{ax}   \text{as second function}  `
\[I = \text{ cos }\left( bx \right)\frac{e^{ax}}{a} - \int - \text{ sin }\left( bx \right) \times b \times \frac{e^{ax}}{a}dx\]
\[ \Rightarrow I = \frac{e^{ax} \text{ cos} \left( bx \right)}{a} + \frac{b}{a}\int\text{ sin } \left( bx \right) e^{ax} dx\]
\[ \Rightarrow I = \frac{e^{ax}}{a}\text{ cos }\left( bx \right) + \frac{b}{a}\int e^{ax} \times \text{ sin } \left( bx \right)dx\]
\[ \Rightarrow I = \frac{e^{ax}}{a}\text{ cos }\left( bx \right) + \frac{b}{a} I_1 . . . . . \left( 1 \right)\]
`  \text{ where I}_1 = \int        e^{ax}    \sin  ( bx )dx `
` \text{ Now, I}_1 = \int   e^{ax} \sin    ( bx )dx  `
` \text{Considering sin ( bx ) as first function }` `\text{ e}^{ax}  \text{ as second function } `
\[ I_1 = \text{ sin } \left( bx \right)\frac{e^{ax}}{a} - \int\text{ cos }\left( bx \right)b\frac{e^{ax}}{a}dx\]
\[ \Rightarrow I_1 = \frac{\text{ sin } \left( bx \right) e^{ax}}{a} - \frac{b}{a}\int e^{ax} \text{ cos} \left( bx \right)dx\]
\[ \Rightarrow I_1 = \frac{e^{ax} \text{ sin } \left( bx \right)}{a} - \frac{b}{a}I . . . . . \left( 2 \right)\]
\[\text{ From ( 1 ) and ( 2)}\]
\[I = \frac{e^{ax}}{a}\text{ cos } \left( bx \right) + \frac{b}{a}\left[ \frac{e^{ax} \text{ sin } \left( bx \right)}{a} - \frac{b}{a}I \right]\]
\[ \Rightarrow I = \frac{e^{ax} \text{ cos }\left( bx \right)}{a} + \frac{b e^{ax} \text{ sin } \left( bx \right)}{a^2} - \frac{b^2}{a^2}I\]
\[ \Rightarrow I\left( 1 + \frac{b^2}{a^2} \right) = e^{ax} \left[ \frac{a \text{ cos } \left( bx \right) + b \text{ sin } \left( bx \right)}{a^2} \right]\]
\[ \therefore I = \frac{e^{ax} \left[ a \text{ cos bx + b }\text{ sin }\left( bx \right) \right]}{\left( a^2 + b^2 \right)} + C\]