Question

\[\int e^{ax} \text{ sin} \left( bx + C \right) dx\]

Answer 1

\[\text{ Let I } = \int e^{ax} \sin \left( bx + C \right)\text{ dx }\]
\[\text{ Considering sin}\ \left( bx + C \right) as\text{ first    function  and }e^{ax}\text{ as second  function}\]
\[I = \text{ sin }\left( bx + C \right)\frac{e^{ax}}{a} - \int \text{ cos }\left( bx + C \right)b\frac{e^{ax}}{a}dx\]
\[ \Rightarrow I = \frac{e^{ax} \text{ sin }\left( bx + C \right)}{a} - \frac{b}{a}\int e^{ax} \text{ cos} \left( bx + C \right) dx\]
\[ \Rightarrow I = \frac{e^{ax} \text{ sin }\left( bx + C \right)}{a} - \frac{b}{a} I_1 . . . \left( 1 \right)\]
\[\text{ where I}_1 = \int e^{ax} \text{ cos } \left( bx + C \right)dx\]
\[\text{ Now, } I_1 = \int e^{ax} \cos \left( bx + C \right)dx\]
\[\text{ Consider  cos}\ \left( bx + C \right)\text{ as first function }e^{ax}\text{ as  second funciton }\]
\[ I_1 = \cos \left( bx + C \right)\frac{e^{ax}}{a} - \int - \sin \left( bx + C \right)b\frac{e^{ax}}{a}dx\]
\[ \Rightarrow I_1 = \frac{e^{ax} \cos \left( bx + C \right)}{a} + \frac{b}{a}\int e^{ax} \sin \left( bx + C \right)dx\]
\[ \Rightarrow I_1 = \frac{e^{ax} \cos \left( bx + C \right)}{a} + \frac{b}{a}I . . . . . \left( 2 \right)\]
` \text{ From ( 1 ) and ( 2 )} `
\[I = \frac{e^{ax} \sin \left( bx + C \right)}{a} - \frac{b}{a}\left[ \frac{e^{ax} \cos \left( bx + C \right)}{a} + \frac{b}{a}I \right]\]
\[ \Rightarrow I = \frac{e^{ax} \sin \left( bx + C \right)}{a} - \frac{b}{a^2} e^{ax} \text{ cos }\left( bx + C \right) - \frac{b^2}{a^2}I\]
\[ \Rightarrow I\left( 1 + \frac{b^2}{a^2} \right) = \frac{e^{ax} \text{ a }\text{ sin }\left( bx + C \right) - b e^{ax} \text{ cos }\left( bx + C \right)}{a^2} + C_1 \]
\[ \Rightarrow I = e^{ax} \frac{\left[ a \text{ sin }\left( bx + C \right) - b \text{ cos} \left( bx + C \right) \right]}{a^2 + b^2} + C_1 \]
`  \text{ Where   C}_{1 } `  `\text{ is  integration  constant } `