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

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

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

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

    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}

    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}

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

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

    Where did I go wrong?
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  2. #2
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    \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
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  3. #3
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    Quote Originally Posted by chengbin View Post
    \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)
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  4. #4
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    Hello,

    Another method :

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


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

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

    Hence
    \begin{aligned}<br />
\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 \\<br />
&=e^{-3x} \cdot \frac{3\cos(x)-\sin(x)}{-10}+C \\<br />
&=\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
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  5. #5
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    How do you box your answer?
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  6. #6
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    \boxed{ CONTENT }

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