Name of this identity? $\int e^{\alpha x}\cos(\beta x)  \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$ Again:
$$\int e^{\alpha x}\cos(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$$
Also the one for $\sin$:
$$\int e^{\alpha x}\sin(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \sin(\beta x)-\beta \cos(\beta x))}{\alpha^2+\beta^2}$$
Both help avoid integration by parts. I find these two proofs very useful and I'd like to know what their names are.
 A: I answered this in another question, but I think it might apply better here:

In general, if you want to find $$ \int e^{ax}\cdot \sin{bx}\cdot dx$$ you can argue as follows:
Note that for any $\alpha$ or $\beta$, you have
$$\eqalign{
  & \frac{d}{{dx}}\left( {{e^{\alpha x}}\sin \beta x} \right) = \alpha {e^{\alpha x}}\sin \beta x + \beta {e^{\alpha x}}\cos \beta x  \cr 
  & \frac{d}{{dx}}\left( {{e^{\alpha x}}\cos \beta x} \right) = \alpha {e^{\alpha x}}\cos \beta x - \beta {e^{\alpha x}}\sin \beta x \cr} $$
so that any integral of the form 
$$ \int e^{\alpha x}\cdot \sin{\beta x}\cdot dx$$ 
is a linear combination of the former functions. Let's then find $c_1$ and $c_2$ such that
$$\frac{d}{{dx}}\left( {{c_1}{e^{\alpha x}}\sin \beta x + {c_2}{e^{\alpha x}}\cos \beta x} \right) = {e^{\alpha x}}\sin \beta x$$
$${c_1}\alpha {e^{\alpha x}}\sin \beta x + {c_1}\beta {e^{\alpha x}}\cos \beta x + {c_2}\alpha {e^{\alpha x}}\cos \beta x - {c_2}\beta {e^{\alpha x}}\sin \beta x = {e^{\alpha x}}\sin \beta x$$
This means we need
$$\eqalign{
  & {c_1}\alpha  - {c_2}\beta  = 1  \cr 
  & {c_1}\beta  + {c_2}\alpha  = 0 \cr} $$
This will yield with little work
$$\eqalign{
  & {c_1} = \frac{\alpha }{{{\alpha ^2} + {\beta ^2}}}  \cr 
  & {c_2} =  - \frac{\beta }{{{\alpha ^2} + {\beta ^2}}} \cr} $$
which means that, in general:
$$\int {{e^{\alpha x}}} \cdot\sin \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \sin \beta x - \beta \cos \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$
Analogously, you will get that
$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

This has no formal name, it is just a formula for finding a primitive.
A: They are the real and imaginary parts of the following identity:
$$
\int{e^{(\alpha+i\beta)x}dx} = \frac{e^{(\alpha+i\beta)x}}{\alpha+i\beta} = \frac{(\alpha - i\beta)e^{(\alpha+i\beta)x}}{\alpha^2 + \beta^2}.
$$
A: HINT
Generally, if we denote $A=\int e^{a x} \cdot \cos (b x) \ dx$ and $B=\int e^{a x} \cdot \sin (b x) \ dx$ and integrate by parts A and B, then we obtain an easy-to-solve system in $A$ and $B$. But this is just another possible approach. However, mjqxxxx's approach is by far a better one.
Chris.
