<s>[INST] Finding Integral of $y = e^{ax}\sin (bx)$ I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above). 
I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or any alternate method available) 
 [/INST] $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large \textbf{Method -1} $ 
$$
\begin{align}
I & = \sin (bx) \left( \cfrac{1}{a} e^{ax}  \right)  - {\int} \left( b\cos (bx) \left(\cfrac{e^{ax}}{a} \right) \right) \\
& = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a} \left[ (\cos (bx)) \cfrac{e^{ax}}{a} + b\sin (bx) \cfrac{e^{ax}}{a} \right] \\
&= \sin (bx) \left( \cfrac{1}{a} (e^{ax}) \right) - \cfrac{b}{a} \left[ \cfrac{1}{a} (\cos (bx) ^{e^{ax}})  +  \cfrac{b}{a} (I) \right]  \\
& = \sin (bx) \left(\cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a^2} \cos (bx) e^{ax} -  \cfrac{b^2}{a^2} (I) \end{align}$$
Now,
$$\begin{align}
I  + \cfrac{b^2}{a^2} (I) & = \sin (bx) \cfrac{1}{a} (e^{ax}) - \cfrac{b}{a^2} (\cos(bx)) e^{ax} \\
I \left(\cfrac{a^2+b^2}{a^2}\right) & = e^{ax} \left(\cfrac{\sin (bx)}{a} - \cfrac{b}{a^2} \cos(bx) \right) \\
I \left(\cfrac{a^2+b^2}{\require{cancel}\cancel{a^2}}\right)& = \cfrac{e^{ax}}{\require{cancel}\cancel{a^2}} \left[a \sin (bx) – b\cos (bx) \right] \\
I (a^2 + b^2) & = e^{ax} \left[ a\sin bx - b \cos bx \right] \end{align}$$
Thus, we have our answer:
$$\begin{align}
I  &= \cfrac{e^{ax}}{a^2 + b^2} \left[a\sin bx - b \cos bx \right]  + \color{grey}{\rm C} \end{align}$$
</s>