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

• May 2nd 2009, 04:46 PM
Fallen186
Integral of e^(-3x)*cos(x) dx
Problem:
$\int e^{-3x}*cos(x) dx$

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

Quote:

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

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

Quote:

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

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

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

Put everything back together

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

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

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

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

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

Where did I go wrong?
• May 2nd 2009, 05:12 PM
chengbin
$\int sin(x)*e^{-3x} = \frac{sin(x)*e^{-3x}}{-3} - \int \frac{cos(x)*e^{-3x}}{-3}$

Integrating sinx becomes -cosx, not cosx.

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

$-\frac{e^{-3x}*(sin(x)+3cos(x))}{10}+C$
• May 2nd 2009, 05:17 PM
Fallen186
Quote:

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

Integrating sinx becomes -cosx, not cosx.

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

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

Thanks, that was my problem(even though it was a derivative that didn't change into a negative)
• May 3rd 2009, 01:53 AM
Moo
Hello,

Another method :

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

\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
\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
\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 :D

In the first quote :
Quote:

u = cos(x)
du = sin(x) dx
This is false, it's du=-sin(x) dx
• May 3rd 2009, 11:58 AM
Fallen186