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Math Help - integral with e^2x

  1. #1
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    integral with e^2x

    Hi,

    I'm trying to solve the integral e^2xsin3x dx. I think I need to use the forumula int. usinudu=sinu-ucosu+c. Is that correct?

    Would u be e^2x
    du=e^2x(2)?????

    So...int. usinudu=sinu-ucosu+c
    =e^2xsine^2x=sine^2x-e^2xcose^2x+c

    I don't think this is correct. Could someone please give me a hint?

    Thank you
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle I=\int e^{2x}\sin 3xdx=\int\left(\frac{e^{2x}}{2}\right)'\sin 3xdx=
    \displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{2}\int e^{2x}\cos 3xdx=
    \displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{2}\int\left(\frac{e^{2x}}{2}\right)'\cos 3xdx=
    \displaystyle =\frac{1}{2}e^{2x}\sin 3x-\frac{3}{4}e^2x\cos 3x-\frac{9}{4}I
    Now, finish it.
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  3. #3
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    See also here.
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  4. #4
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    Hi,





    Now, finish it.
    Can you please explain to me what the I's stand for and why you put '? Is the ' necessary? Can you explain what you did? I don't really understand it. We just started doing these types of problems.
    Thank you very much
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  5. #5
    MHF Contributor red_dog's Avatar
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    I represents the initial integral.
    I used integration by parts:
    \int f'(x)g(x)dx=f(x)g(x)-\int f(x)g'(x)dx
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  6. #6
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    See the link I gave you.

    You can derive a general formula using integration by parts.
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  7. #7
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    Can one do it with a forumla? We're supposed to be using formulas now.

    Wouldn't int. usindu=sinu-ucosu+c work or do you have to do integration by parts?

    Thank you
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  8. #8
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    \int x\sin\,dx & \int e^x\sin x\,dx are two different integrals.

    Use integration by parts.
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