# Thread: Integral of e^(-3x)*cos(x) dx

1. ## Integral of e^(-3x)*cos(x) dx

Problem:
$\displaystyle \int e^{-3x}*cos(x) dx$

Attempt:
$\displaystyle \int uv' = uv - \int u'v$

u = $\displaystyle cos(x)$
du = $\displaystyle sin(x) dx$
v = $\displaystyle \frac{e^{-3x}}{-3}$
dv = $\displaystyle e^{-3x}$
$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{1}{3} \int sin(x)*e^{-3x}$

Focusing on $\displaystyle \int sin(x)*e^{-3x}$

u = $\displaystyle sin(x)$
du = $\displaystyle cos(x) dx$
v = $\displaystyle \frac{e^{-3x}}{-3}$
dv = $\displaystyle e^{-3x}$
$\displaystyle \int sin(x)*e^{-3x} = \frac{sin(x)*e^{-3x}}{-3} - \int \frac{cos(x)*e^{-3x}}{-3}$

Focusing on $\displaystyle - \int \frac{cos(x)*e^{-3x}}{-3}$

$\displaystyle \frac {1}{3} \int cos(x)*e^{-3x}$

Put everything back together

$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{1}{3}[\frac{sin(x)*e^{-3x}}{-3} + \frac {1}{3} \int cos(x)*e^{-3x}]$

$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{sin(x)*e^{-3x}}{-9} + \frac {1}{9} \int cos(x)*e^{-3x}]$

$\displaystyle \frac{8}{9}\int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{sin(x)*e^{-3x}}{-9}$

$\displaystyle \int e^{-3x}*cos(x) dx= \frac{3*cos(x)*e^{-3x}}{-8} + \frac{sin(x)*e^{-3x}}{-8}+C$

$\displaystyle \frac{e^{-3x}*(sin(x)-3cos(x))}{10}+C$

Where did I go wrong?

2. $\displaystyle \int sin(x)*e^{-3x} = \frac{sin(x)*e^{-3x}}{-3} - \int \frac{cos(x)*e^{-3x}}{-3}$

Integrating sinx becomes -cosx, not cosx.

I tried this problem, and I made a mistake somewhere because I got

$\displaystyle -\frac{e^{-3x}*(sin(x)+3cos(x))}{10}+C$

3. Originally Posted by chengbin
$\displaystyle \int sin(x)*e^{-3x} = \frac{sin(x)*e^{-3x}}{-3} - \int \frac{cos(x)*e^{-3x}}{-3}$

Integrating sinx becomes -cosx, not cosx.

I tried this problem, and I made a mistake somewhere because I got

$\displaystyle -\frac{e^{-3x}*(sin(x)+3cos(x))}{10}+C$
Thanks, that was my problem(even though it was a derivative that didn't change into a negative)

4. Hello,

Another method :

$\displaystyle \cos(x)=\text{Re}(e^{ix})$

\displaystyle \begin{aligned} \Rightarrow \int e^{-3x} \cos(x) ~dx &=\text{Re}\left(\int e^{-3x}e^{ix} ~dx\right) \\ &=\text{Re}\left(\int e^{(i-3)x} ~dx\right) \\ &=\text{Re}\left(\frac{e^{(i-3)x}}{i-3}\right) +C\end{aligned}

And
\displaystyle \begin{aligned} \frac{e^{(i-3)x}}{i-3} &=e^{-3x} \cdot \frac{e^{ix}}{i-3} \cdot \frac{i+3}{i+3} \\ &=e^{-3x}\cdot \frac{(\cos(x)+i\sin(x)) \cdot (i+3)}{-1-9} \\ &=e^{-3x} \cdot \frac{[3\cos(x)-\sin(x)]+i[\cos(x)+3\sin(x)]}{-10} \end{aligned}

Hence
\displaystyle \begin{aligned} \int e^{-3x} \cos(x) ~dx&=\text{Re}\left(e^{-3x} \cdot \frac{[3\cos(x)-\sin(x)]+i[\cos(x)+3\sin(x)]}{-10}\right)+C \\ &=e^{-3x} \cdot \frac{3\cos(x)-\sin(x)}{-10}+C \\ &=\boxed{e^{-3x}\cdot \frac{\sin(x)-3\cos(x)}{10}}+C\end{aligned}

What's good with this is that there is no need to integrate cos or sin, and mess up with the signs

In the first quote :
u = cos(x)
du = sin(x) dx
This is false, it's du=-sin(x) dx

6. \boxed{ CONTENT }

You can always quote a post to take a look at its contents (or click the latex image).

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