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Math Help - integrate [E^(2x)](Cos[2x])

  1. #1
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    integrate [E^(2x)](Cos[2x])

    please help integrate [E^(2x)](Cos[2x])
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by guess
    please help integrate [E^(2x)](Cos[2x])
    <br />
e^{2x} \cos(2x)= \mbox{Re}\ (e^{2x} e^{2x \mathrm{i}})=\mbox{Re}\ (e^{2x(1+\mathrm{i})})<br />

    Hence:

    <br />
\int e^{2x} \cos(2x) dx= \int \mbox{Re}\ (e^{2x(1+\mathrm{i})})dx=\mbox{Re}\ \left(\int e^{2x(1+\mathrm{i})}dx \right)<br />

    <br />
\int e^{2x} \cos(2x) dx= \mbox{Re}\ \left(\frac{e^{2x(1+\mathrm{i})}}{2(1+\mathrm{i})}  \right) + C<br />
.

    You should be able to complete the simplification from here.

    RonL
    Last edited by CaptainBlack; April 12th 2006 at 11:46 AM.
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  3. #3
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    Quote Originally Posted by guess
    please help integrate [E^(2x)](Cos[2x])
    You have,
    \int e^{2x}\cos 2xdx
    Integrate by parts, u'=e^{2x}\mbox{ and } v=\cos 2x
    \frac{1}{2}e^{2x}\cos 2x+\int e^{2x}\sin 2x dx
    Integrate by parts, u'=e^{2x}\mbox{ and } v=\sin 2x
    \frac{1}{2}e^{2x}\cos 2x+\left( \frac{1}{2}e^{2x}\sin 2x-\int e^{2x}\cos 2x dx \right).
    Thus,
    \int e^{2x}\cos 2xdx=\frac{1}{2}e^{2x}\cos 2x+\left( \frac{1}{2}e^{2x}\sin 2x-\int e^{2x}\cos 2x dx \right)
    Thus, (combine integrals),
    2\int e^{2x}\cos 2xdx=\frac{1}{2}e^{2x}\cos 2x+ \frac{1}{2}e^{2x}\sin 2x
    Thus,
    \int e^{2x}\cos 2xdx=\frac{1}{4}e^{2x}(\sin 2x+\cos 2x)+C
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