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    How on earth can I integrate this:

    -x*(e^x)(sin(x))

    ?
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    first find \int e^xsinx\ dx then do IBP on \int -xe^xsinx\ dx with u=-x and dv/dx = e^xsinx
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    (Original post by Pheylan)
    first find \int e^xsinx\ dx then do IBP on \int -xe^xsinx\ dx with u=-x and dv/dx = e^xsinx
    I got \frac{1}{2}e^x(sinx - cosx) for first bit

    Is this anywhere near right?
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    (Original post by Dagnabbit)
    I got \frac{1}{2}e^x(sinx - cosx) for first bit

    Is this anywhere near right?
    if memory serves, yes
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    (Original post by lilangel890)
    Do you have to do it by parts? If not you can do it by log differentiation and it's much simpler:

    Let y=-xe^xsinx

    Log both sides. We know ln(abc) = lna + lnb + lnc. Use this to split the expression and then differentiate wrt x (remembering left hand side is implicit, so you get 1/y dy/dx)

    Once you multiply up by the y and put in a nicer form you will get the required answer
    gosh, that's so much simpler. Cheers!
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    (Original post by Dagnabbit)
    gosh, that's so much simpler. Cheers!
    it's also incorrect
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    :mad:
    (Original post by Pheylan)
    it's also incorrect
    Get your facts straight nub. I did it and checked it in Matlab and it's right
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    (Original post by lilangel890)
    :mad:

    Get your facts straight nub. I did it and checked it in Matlab and it's right
    what answer did you get?
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    (Original post by lilangel890)
    :mad:

    Get your facts straight nub. I did it and checked it in Matlab and it's right
    What on Earth are you on about? You've just differentiated it, whereas the OP is trying to integrate... Pheylan's advice is correct.
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    (Original post by lilangel890)
    :mad:

    Get your facts straight nub. I did it and checked it in Matlab and it's right
    Yes but you've invented your own question.
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    While on integration

    integral of

    e^(-x) * sine sqared x
    ???
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    (Original post by ck_no_1)
    While on integration

    integral of

    e^(-x) * sine sqared x
    ???
    It's a tricky one. You need to use IBP 3 times and also be able to spot the trick for integrating things of the form e^{kx}\sin (nx). The answer isn't that messy.
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    for the complex conscious... or if you just hate using paper
    -\int xe^{(1+i)x}\mbox{d}x=\frac{e^{(i  +1)x}\left[1-\left(i+1\right)x\right]}{\left(1+i\right)^{2}}+C= \frac{e^{x}}{2}\left(ie^{ix}(1+i  )x-ie^{ix}\right)+C

    taking the imaginary part gives: \frac{e^{x}}{2}\left[\left(\cos x-\sin x\right)x-\cos x\right]+C and we're done.

    all you need to know is that  e^{ix}= \cos x+i \sin x and treat i as a constant when integrating

    For the integral  \int e^{-x}\sin^{2}x\mbox{d}x , try the identity  \sin^{2}x=\frac{1}{2}\left(1-\cos2x\right) and integrate by parts as normally
 
 
 
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