0
$\begingroup$

The following function is being integrated:

$$e^{-3x}\sin(3x)$$

For the ease of time, I have managed to calculate it down to the following formula using integration by parts:

$$0=-e^{-3x}\cos(3x)+e^{-3x}\sin(3x)$$

The above phrase is calculated by splitting the first phrase into two.

Where $$u=e^{-3x}$$ and $$v=\sin(3x)$$

Would this be an appropriate solution in terms of its simplification and finding the integral of the initial equation? Can I go and do anything further?

$\endgroup$
  • 1
    $\begingroup$ I am not sure about how you got your formula, but try doing the integration by parts twice. Take $u=e^{-3x}$ and $dv=\sin(3x)$ for example. The new term will have exponential and cosine in it, again take $u$ to be the exponential part and $dv$ to be the cosine part of the integral. Finally you'll get the first term you started with, gather them on one side, and what's left on the other side will be the answer $\endgroup$ – Fareed Abi Farraj Jul 22 '19 at 8:43
  • $\begingroup$ You can even do it simpler without integration by parts (but this is not what you want). $\endgroup$ – Claude Leibovici Jul 22 '19 at 8:46
  • 1
    $\begingroup$ The formula $$0=-e^{-3x}cos(3x)+e^{-3x}sin(3x)$$ is incorrect. For example, it is false for $x=0$. Did you mean $$0=-\int e^{-3x}cos(3x)dx+\int e^{-3x}sin(3x)dx?$$ $\endgroup$ – 5xum Jul 22 '19 at 8:48
  • $\begingroup$ @Claude Leibovici may you show us your method please ? Is it the same as the method of coreyman317 or a new one ? $\endgroup$ – Fareed Abi Farraj Jul 22 '19 at 20:57
  • 1
    $\begingroup$ @FareedAF. Exactly the same (using complex). $\endgroup$ – Claude Leibovici Jul 23 '19 at 2:34
3
$\begingroup$

In general the formula for integration by parts is $\int uv'=uv-\int u'v$.

If $u = e^{-3x}$, then $u'=-3e^{-3x}$ and if $v' = \sin(3x)$, then $v = -\frac{1}{3}\cos(3x)$.

So $\int e^{-3x} \sin(3x)dx = -\frac{1}{3}e^{-3x}\cos(3x)-\color{red}{\int e^{-3x} \cos(3x)dx} + C$.

If $u = e^{-3x}$, then $u'=-3e^{-3x}$ and if $v' = \cos(3x)$, then $v = \frac{1}{3}\sin(3x)$.

So $\color{red}{\int e^{-3x} \cos(3x)dx} = \color{blue}{\frac{1}{3}e^{-3x}\sin(3x)+\int e^{-3x}\sin(3x)} + C'$.

Plug in the result for the red part in the equation above and you get:

$\int e^{-3x} \sin(3x)dx = -\frac{1}{3}e^{-3x}\cos(3x)-\left(\color{blue}{\frac{1}{3}e^{-3x}\sin(3x)+\int e^{-3x}\sin(3x)}\right)+C''$

Solve this equation for $\int e^{-3x} \sin(3x)dx$ and you get this:

$\int e^{-3x} \sin(3x)dx = -\frac{1}{6}e^{-3x}(\cos(3x)+\sin(3x))+C''$

$\endgroup$
4
$\begingroup$

Here's a fun way to do it if you're familiar with complex numbers: Let$$I=\int e^{-3x}\sin(3x)dx=\int e^{-3x}\Im e^{3ix}dx=\tag1$$ $$\Im\int e^{-3x+3ix}dx=\Im \int e^{x(-3+3i)}dx=\Im\left(\frac{1}{-3+3i}e^{-3x+3ix}\right)\tag2$$ $$=\Im\left(e^{-3x}\left(-\frac{1}{6}-\frac{i}{6}\right)\left(\cos(3x)+i\sin(3x)\right)\right)\tag3$$ $$=-\frac{e^{-3x}}{6}\Im\left(-\sin(3x)+\cos(3x)+i\left(\sin(3x)+\cos(3x)\right)\right)\tag4$$ $$\boxed{=-\frac{e^{-3x}}{6}(\sin(3x)+\cos(3x))+C}\tag5$$

Alternatively, integrate by parts twice making sure to keep track of your integral by naming it the 'variable' $I$; namely

Let $$I=\int e^{-3x}\sin(3x)dx$$

Integrate by parts, $$u=e^{-3x}\implies du=-3e^{-3x}\space dx$$ $$dv=\sin(3x)dx\implies v=-\frac{1}{3}\cos(3x)$$ Then $$I=-\frac{1}{3} e^{-3x}\cos(3x)-\int e^{-3x}\cos(3x)dx $$

Integrate by parts one more time on the last integral, $$t=e^{-3x}\implies dt=-3e^{-3x}\space dx$$ $$dw=\cos(3x)dx\implies w=\frac{1}{3}\sin(3x)$$

so that $$I=-\frac{1}{3} e^{-3x}\cos(3x)-\left(\frac{1}{3}e^{-3x}\sin(3x)-\int-e^{-3x}\sin(x)dx\right) $$ $$=-\frac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)-I\implies2I=-\frac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)$$ So dividing by $2$ on both sides gives $$I=-\frac{e^{-3x}}{6}\left(\sin(3x)+\cos(3x)\right)+C$$ matching the above formula. $$$$

$(1)$: Utilize Euler's formula $e^{3ix}=\cos(3x)+i\sin(3x)$ and $\Im(e^{3ix})=\sin(3x)$

$(2)$: Utilize the facts: $e^ae^b=e^{a+b}$ and $\int e^{kx}dx=\frac{1}{k}e^{kx}+c$

$(3)$: Simplify $\frac{1}{-3+3i}$ and expand $e^{-3x+3ix}=e^{-3x}e^{3ix}=e^{-3x}(\cos(3x)+i\sin(3x))$

$(4)$: Expand $\left(-\frac{1}{6}-\frac{i}{6}\right)\left(\cos(3x)+i\sin(3x)\right)$ and collect real and imaginary parts

$(5)$: Take the imaginary part

$\endgroup$
1
$\begingroup$

You can do this way using two integrations by part. The idea is to integrate the $e^{-3x}$ part twice and to recognize the original integral after the second integration by part.

\begin{align*}I&=\int e^{-3x}\sin(3x) dx=\left[-\dfrac{e^{-3x}}{3} \sin(3x)\right]+\int_a^b {e^{-3x}}\cos(3x)dx\\ &=\left[-\dfrac{e^{-3x}}{3} \sin(3x)\right] + \left[-\dfrac{e^{-3x}}{3} \cos(3x)\right] - \int {e^{-3x}}\sin(3x)dx\\ 2I&=-\dfrac{e^{-3x}}{3}\left(\sin(3x)+\cos(3x)\right)+C \end{align*} Hence $$I=-\dfrac{e^{-3x}}{6}\left(\sin(3x)+\cos(3x)\right)+C.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.