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) 

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad  \large{\textbf{Method - 2}}$
Let : $I = \int e^{ax} \sin (bx) $
And say : 
$$y_1  = e^{ax} \sin (bx) \\
y_2 = e^{ax} \cos (bx) $$
And : $$y'_1 = ae^{ax}\sin (bx) + e^{ax}b\cos (bx) \\
y'_2 = -e^{ax} b\sin (bx) + ae^{ax} \cos (bx) $$
Or we can also write $y'_1$ and $y'_2$ in the form of $y_1$ and $y_2$ as: $$y'_1 = ay_1 + by_2 \\ y'_2 = -by_1 + ay_2 $$ 
In a matrix, this can be written as : $$\bar y ' = \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] \bar y$$ 
Inverting the Derivative Matrix :
$$\begin{align} D^{-1} = \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a & b \\ -b & a \end{matrix} \right] \\
= \left[ \begin{matrix} \cfrac{a}{a^2+b^2} & \cfrac{b}{a^2 + b^2} \\ \cfrac{-b}{a^2 + b^2} & \cfrac{a}{a^2 + b^2} \end{matrix} \right] \times \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\
= \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a \\ -b \end{matrix} \right] \\
I = \cfrac{e^{ax}}{a^2+b^2}  (a \sin bx - b \cos bx ) + C \end{align}$$
