Question

Write a value of\[\int e^{ax} \sin\ bx\ dx\]

Answer 1

\[\text{ Let I }= \int e^{ax} . \sin bx\ dx\]
\[ = \sin bx\int e^{ax}\text{  dx }- \int\left\{ \frac{d}{dx}\left( \sin bx \right)\int e^{ax} dx \right\}dx\]
\[ = \sin bx \times \frac{e^{ax}}{a} - \int\cos bx \times b . \frac{e^{ax}}{a}\]
\[ = \sin bx \times \frac{e^{ax}}{a} - \frac{b}{a}\int e^{ax} . \cos bx\ dx \]
\[ = \sin bx \times \frac{e^{ax}}{a} - \frac{b}{a} I_1 . . . \left( 1 \right)\]
\[ \therefore I_1 = \int e^{ax} \times \cos bxdx\]
\[ = \cos bx\int e^{ax} dx - \int\left\{ \frac{d}{dx}\left( \cos bx \right)\int e^{ax} dx \right\}dx\]
\[ = \cos bx \times \frac{e^{ax}}{a} + \int  b . \sin bx \times \frac{e^{ax}}{a}dx\]
\[ = \cos bx . \frac{e^{ax}}{a} + \frac{b}{a}I . . . . \left( 2 \right)\]
\[\text{ From }\left( 1 \right) \text{ and}\ \left( 2 \right)\]
\[ \therefore I = \sin bx . \frac{e^{ax}}{a} - \frac{b}{a} \left\{ \cos bx . \frac{e^{ax}}{a} + \frac{b}{a}I \right\}\]
\[ \Rightarrow I = \sin bx . \frac{e^{ax}}{a} - \frac{b}{a^2} \cos bx \text{ e}^{ax} - \frac{b^2}{a^2}I\]
\[ \Rightarrow I + \frac{b^2}{a^2}I = \sin bx . \frac{e^{ax}}{a} - \frac{b \cos bx \text{ e}^{ax}}{a^2}\]
\[ \Rightarrow \left( a^2 + b^2 \right)I = \left( a \sin bx - b\cos bx \right) e^{ax} \]
\[ \Rightarrow I = \frac{\left( a \sin bx - b\cos bx \right) e^{ax}}{a^2 + b^2} + C\]