Finding $\int e^{2x} \sin{4x} \, dx$ Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$
Highlighted in red, I seem to integrating an exponential times trig expression again ... doesn't seem like te right way to go
So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) - \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?
Am I using the wrong substitution?
 A: 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
$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$
Hope this helps!
A: Take your red integral and integrate it by parts again, making sure to let $u$ be the exponential function again. You will get that
The integral you care about = stuff + the integral you care about*(other stuff)
and therefore
the integral you care about = stuff/(1 - other stuff)
A: I would jump into complex values,
and noting that
$\sin 4x = \Im e^{4 i x}$ 
(I wrote \Im there, wanting to get the imaginary part, and got the $\Im$),
$$\int e^{2x} \sin 4x \, dx
= \Im \int e^{2x+4ix} \, dx
= \Im \int e^{x(2+4i)} \, dx
= \Im \frac{e^{x(2+4i)}}{2+4i}.
$$
Since $\frac{1}{2+4i} = \frac{2-4i}{20} = \frac{1-2i}{10}$
and $e^{x(2+4i)} = e^{2x}(\cos 4x + i \sin 4x)$,
I get, disregarding the $\frac{e^{2x}}{10}$ for a moment,
$\Im (1-2i)(\cos 4x + i \sin 4x)
= \Im ((\cos 4x + 2 \sin 4x) + i(-2\cos 4x + \sin 4x))
$
$
= -2\cos 4x + \sin 4x
$
so my final result is
$\frac{e^{2x}(-2\cos 4x + \sin 4x)}{10}$.
As a check the derivative of
$e^{2x}(-2\cos 4x + \sin 4x)$ (disregarding the 1/10 for now) is
$$
\begin{align}
& 2e^{2x}(-2\cos 4x + \sin 4x) + e^{2x}(8\sin 4x + 4\cos 4x) \\[4pt]
= {} & e^{2x}(-4\cos 4x + 2\sin 4x + 8\sin 4x + 4\cos 4x)
= e^{2x} (10 \sin 4x).
\end{align}
$$
A: I wonder how many times this same question has been asked here, and how many times it is asked every semester in every second-semester calculus course?
If you integrate by parts a second time, you get the same integral you started with, and the naive reaction is that that means you're getting nowhere.  But in truth it means you're almost done.  You've got
$$
\int \text{something} = \text{something} - \text{something}\cdot \int [\text{same integral}].
$$
So you add the same thing to both sides of the equation and get
$$
\int \text{something} + \text{something}\cdot\int [\text{same integral}] = \text{something} + C
$$
Then you write
$$
(1+\text{something}) \cdot\int\cdots\cdots = \text{something} + C
$$
$$
\int\cdots\cdots = \frac{\text{something}}{1+\text{something}} + \text{constant}.
$$
A: To find $\int e^{2x}\sin4x\,dx,$ we can let $u=e^{2x}$ and $dv=\sin4x\,dx,$ which implies $du=2e^{2x}\,dx$ and $v=-\frac{1}{4}\cos4x.$ Therefore, $$\begin{align}\int e^{2x}\sin4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(-\frac{1}{4}\cos4x\right)-\int-\frac{1}{4}\cos4x\left(2e^{2x}\right)\,dx\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\int e^{2x}\cos4x\,dx.\end{align}$$ Apply another integration for $\int e^{2x}\cos4x\,dx,$ this time letting $u=e^{2x}$ and $dv=\cos4x\,dx,$ so that $du=2e^{2x}\,dx$ and $v=\frac{1}{4}\sin4x.$ Hence, $$\begin{align}\int e^{2x}\cos4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(\frac{1}{4}\sin4x\right)-\int\frac{1}{4}\sin4x\left(2e^{2x}\right)\,dx\\ &= \frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx.\end{align}$$
Combining both formulas, we get $$\begin{align}\int e^{2x}\sin4x\,dx &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\left(\frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx\right)\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x-\frac{1}{4}\int e^{2x}\sin4x\,dx.\end{align}$$ So, $$\frac{5}{4}\int e^{2x}\sin4x\,dx=-\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x.$$ Therefore, multiplying by $\frac{4}{5}$ and adding in the constant of integration, we see that $$\int e^{2x}\sin4x\,dx=-\frac{1}{5}e^{2x}\cos4x+\frac{1}{10}e^{2x}\sin4x+C,$$ which we can rewrite as $$\int e^{2x}\sin4x\,dx=\frac{1}{10}e^{2x}\left(-2\cos4x+\sin4x\right)+C.$$
