Problem:
$\displaystyle \int e^{-3x}*cos(x) dx$
Attempt:
$\displaystyle \int uv' = uv - \int u'v $
$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{1}{3} \int sin(x)*e^{-3x} $u = $\displaystyle cos(x)$
du = $\displaystyle sin(x) dx $
v = $\displaystyle \frac{e^{-3x}}{-3}$
dv = $\displaystyle e^{-3x} $
Focusing on $\displaystyle \int sin(x)*e^{-3x} $
$\displaystyle \int sin(x)*e^{-3x} = \frac{sin(x)*e^{-3x}}{-3} - \int \frac{cos(x)*e^{-3x}}{-3}$u = $\displaystyle sin(x)$
du = $\displaystyle cos(x) dx $
v = $\displaystyle \frac{e^{-3x}}{-3}$
dv = $\displaystyle e^{-3x} $
Focusing on $\displaystyle - \int \frac{cos(x)*e^{-3x}}{-3}$
$\displaystyle \frac {1}{3} \int cos(x)*e^{-3x}$
Put everything back together
$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{1}{3}[\frac{sin(x)*e^{-3x}}{-3} + \frac {1}{3} \int cos(x)*e^{-3x}]$
$\displaystyle \int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{sin(x)*e^{-3x}}{-9} + \frac {1}{9} \int cos(x)*e^{-3x}]$
$\displaystyle \frac{8}{9}\int e^{-3x}*cos(x) dx= \frac{cos(x)*e^{-3x}}{-3} + \frac{sin(x)*e^{-3x}}{-9}$
Answer:
$\displaystyle \int e^{-3x}*cos(x) dx= \frac{3*cos(x)*e^{-3x}}{-8} + \frac{sin(x)*e^{-3x}}{-8}+C$
Correct Answer:
$\displaystyle \frac{e^{-3x}*(sin(x)-3cos(x))}{10}+C$
Where did I go wrong?