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Math Help - Integrate

  1. #1
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    Integrate

    \int e^x \sin 2x dx
    I used the integration by parts method.
    \int v\frac{du}{dx}=(uv)-\int u\frac{dv}{dx}
    I let  v=\sin 2x\rightarrow \frac{dv}{dx}=2\cos 2x
    and \frac{du}{dx}=e^x\rightarrow u=e^x
    so,
    uv-\int u\frac{dv}{dx}
    =e^x(\sin 2x)-\int 2e^x\cos 2x dx
    This is the part I am stuck at. If I integrate 2e^x\cos 2x by parts I get another product. And it will continue for I don't know what because I can't get rid of the e^x.
    Any help? I've definitely done something wrong. Thanks
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  2. #2
    Super Member Random Variable's Avatar
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    You'll get the same integral on both sides of the equation.
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  3. #3
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    Quote Originally Posted by arze View Post
    \int e^x \sin 2x dx
    I used the integration by parts method.
    \int v\frac{du}{dx}=(uv)-\int u\frac{dv}{dx}
    I let  v=\sin 2x\rightarrow \frac{dv}{dx}=2\cos 2x
    and \frac{du}{dx}=e^x\rightarrow u=e^x
    so,
    uv-\int u\frac{dv}{dx}
    =e^x(\sin 2x)-\int 2e^x\cos 2x dx
    This is the part I am stuck at. If I integrate 2e^x\cos 2x by parts I get another product. And it will continue for I don't know what because I can't get rid of the e^x.
    Any help? I've definitely done something wrong. Thanks
    Call the original integral (the one you want to calculate) I, and then proceed as you did. Now perform integration by parts on what you arrived at, and you will end up with an integral which is a scalar multiple of the original integral you called I. Rearrange and you will arrive at your answer.

    This is a well known trick for dealing with these, and other similar types of integrals.

    Good luck.
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  4. #4
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    Quote Originally Posted by pomp View Post
    Call the original integral (the one you want to calculate) I, and then proceed as you did. Now perform integration by parts on what you arrived at, and you will end up with an integral which is a scalar multiple of the original integral you called I. Rearrange and you will arrive at your answer.

    This is a well known trick for dealing with these, and other similar types of integrals.

    Good luck.
    Thanks! But I think i haven't learned this trick.
    so
    I=e^x(\sin 2x)-\int 2e^x\cos 2x dx
    then I integrate 2e^x\cos 2x dx?
    Thanks
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  5. #5
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    Quote Originally Posted by arze View Post
    Thanks! But I think i haven't learned this trick.
    so
    I=e^x(\sin 2x)-\int 2e^x\cos 2x dx
    then I integrate 2e^x\cos 2x dx?
    Thanks
    Call the original integral I,

    I = \int e^x \sin{2x} dx
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  6. #6
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    Sorry I don't understand. From my workings, I=e^x(\sin 2x)-\int 2e^x\cos 2x dx? Then what's next?
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  7. #7
    Super Member Random Variable's Avatar
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     \int e^{x} \sin 2x \ dx = e^x \sin 2x - 2 \int e^{x} \cos 2x \ dx = e^x \sin 2x -2e^{x} \cos 2x - 4 \int e^{x} \sin 2x \ dx

    then  5 \int e^{x} \sin 2x \ dx = e^x \sin 2x -2e^{x} \cos 2x

    so  \int e^{x} \sin 2x \ dx = \frac{1}{5}e^{x} \sin 2x - \frac{2}{5} e^{x} \cos 2x + C
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