Question

\[\int e^{- 2x} \sin x\ dx\]

Answer 1

\[\text{ Let I }= \int e^{- 2x} \text{ sin x dx }\]
`\text{Considering sin   x  as first function and` `\text{ e}^{-2x}`   ` \text{ as second function} `
\[I = \sin x\frac{e^{- 2x}}{- 2} - \int\cos x\left( \frac{e^{- 2x}}{- 2} \right)dx\]
\[ \Rightarrow I = - \frac{e^{- 2x} \sin x}{2} + \frac{1}{2}\int e^{- 2x} \text{ cos  x  dx }\]
\[ \Rightarrow I = - \frac{e^{- 2x} \sin x}{2} + \frac{I_1}{2} \text{ where } . . . . . \left( 1 \right)\]
\[\text{ Where,} I_1 = \int e^{- 2x} \text{ cos  x  dx }\]
`\text{Considering cos x as first function and` `\text{ e}^{-2x}`   ` \text{ as second function} `
\[ I_1 = \frac{\text{ cos x  e}^{- 2x}}{- 2} - \int\left( - \sin x \right)\frac{e^{- 2x}}{- 2}dx\]
\[ \Rightarrow I_1 = \frac{e^{- 2x} \cos x}{- 2} - \int\frac{\text{ sin x e}^{- 2x} dx}{2}\]
\[ \Rightarrow I_1 = \frac{- e^{- 2x} \cos x}{2} - \frac{I}{2} . . . . . \left( 2 \right)\]
\[\text{ From }\left( 1 \right) \text{ and }\ \left( 2 \right)\]
\[I = \frac{- e^{- 2x} \sin x}{2} + \frac{1}{2}\left[ \frac{- e^{- 2x} \cos x}{2} - \frac{I}{2} \right]\]
\[ \Rightarrow I + \frac{I}{4} = \frac{- e^{- 2x} \sin x}{2} - \frac{e^{- 2x} \cos x}{4}\]
\[ \Rightarrow \frac{5I}{4} = \frac{- e^{- 2x} \left( 2 \sin x + \cos x \right)}{4}\]
\[ \therefore I = \frac{e^{- 2x}}{5}\left( - 2 \sin x - \cos x \right) + C\]