Question

$\int e^{-2x}\sin xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^{-2x}\sin xdx \\ \mathrm{Considering}\sin x\mathrm{as}\mathrm{first}\mathrm{function}\mathrm{and}{e}^{-2x}\mathrm{as}\mathrm{second}\mathrm{function} \\ I=\sin x\frac{e^{-2x}}{-2}-\int \cos x\left(\frac{e^{-2x}}{-2}\right)dx \\ \Rightarrow I=-\frac{e^{-2x}\sin x}{2}+\frac{1}{2}\int e^{-2x}\cos xdx \\ \Rightarrow I=-\frac{e^{-2x}\sin x}{2}+\frac{I_1}{2}\mathrm{where}.....\left(1\right) \\ \mathrm{Where},I_1=\int e^{-2x}\cos xdx \\ \mathrm{Considering}\cos x\mathrm{as}\mathrm{first}\mathrm{and}{e}^{-2x}\mathrm{as}\mathrm{second}\mathrm{function} \\ {I}_1=\frac{\cos x{e}^{-2x}}{-2}-\int \left(-\sin x\right)\frac{e^{-2x}}{-2}dx \\ \Rightarrow I_1=\frac{e^{-2x}\cos x}{-2}-\int \frac{\sin x{e}^{-2x}dx}{2} \\ \Rightarrow I_1=\frac{-e^{-2x}\cos x}{2}-\frac{I}{2}.....\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right) \\ I=\frac{-e^{-2x}\sin x}{2}+\frac{1}{2}\left[\frac{-e^{-2x}\cos x}{2}-\frac{I}{2}\right] \\ \Rightarrow I+\frac{I}{4}=\frac{-e^{-2x}\sin x}{2}-\frac{e^{-2x}\cos x}{4} \\ \Rightarrow \frac{5I}{4}=\frac{-e^{-2x}\left(2\sin x+\cos x\right)}{4} \\ \therefore I=\frac{e^{-2x}}{5}\left(-2\sin x-\cos x\right)+\mathrm{C} \end{gathered}$$