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Math Help - indefinite integral

  1. #1
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    indefinite integral

    I seem to be getting worse and worse at integrating haha.. I can get this in a form that looks simple but for the life of me cannot seem to get past it!!

     \int \frac{sin(kx)}{\sqrt{2+cos(kx)}}dx where k does not equal zero

    let  cos(kx) = u therefore  dx = \frac{du}{-ksin(kx)}

    so it becomes  -\frac{1}{k} \int \frac{du}{\sqrt{2 + u}}

    for some reason I cannot get past here.. Very frustrating!!
    Thanks in advance,
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by U-God View Post
    I seem to be getting worse and worse at integrating haha.. I can get this in a form that looks simple but for the life of me cannot seem to get past it!!

     \int \frac{sin(kx)}{\sqrt{2+cos(kx)}}dx where k does not equal zero

    let  cos(kx) = u therefore  dx = \frac{du}{-ksin(kx)}

    so it becomes  -\frac{1}{k} \int \frac{du}{\sqrt{2 + u}}

    for some reason I cannot get past here.. Very frustrating!!
    Thanks in advance,
    Try this substitution [for the original integral]:

    Let u=\cos(kx){\color{red}+2}

    See how you go with this one

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Try this substitution [for the original integral]:

    Let u=\cos(kx){\color{red}+2}

    See how you go with this one

    --Chris
    you're a genius
    thanks tons
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  4. #4
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    Actually one more while we're at it would be fantastic!

     \int \frac{1}{5x - \sqrt{x}}dx
    I don't really know how to begin this one, I tried looking for a substitution but couldn't see one that work (although I may be blind).

    Cheers,
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by U-God View Post
    Actually one more while we're at it would be fantastic!

     \int \frac{1}{5x - \sqrt{x}}dx
    I don't really know how to begin this one, I tried looking for a substitution but couldn't see one that work (although I may be blind).

    Cheers,
    Note that \frac{1}{5x-\sqrt{x}}=\frac{1}{\sqrt{x}\left(5\sqrt{x}-1\right)}

    Then make the substitution u=5\sqrt{x}-1

    See how you go with this one

    --Chris
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    I can never see things like you do!! Cheers!
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  7. #7
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    Okay, I got one here that I thought I'd done right but it doesn't match the answer:

     \int \sqrt{1 + 4x^2} dx so I let  x = \frac{1}{2}sinh(u)
    therefore  dx = \frac{1}{2}cosh(u)du


    subbed it in:  \frac{1}{2}\int cosh^2(u)du

    used identity to re-arrange:  \frac{1}{4} \int cosh(2u)du + \frac{1}{4}\int du

    evaluated:  \frac{1}{8}sinh(u) + \frac{1}{4}u + C

    re arranged the substitution I made earlier for  u = arcsinh(2x)
    subbed u back in yielded  \frac{1}{8}sinh(arcsinh(2x)) + \frac{1}{4}arcsinh(2x) + C

    which cancelled to give  \frac{1}{4}x + \frac{1}{4} arcsinh(2x) + C

    however the answer I'm given is:  \frac{1}{4}(arcsinh(2x) + 2x\sqrt{1 + 4x^2}) + C

    Hopefully someone can spot what I've done wrong. Thanks!
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  8. #8
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    Quote Originally Posted by U-God View Post
    Okay, I got one here that I thought I'd done right but it doesn't match the answer:

     \int \sqrt{1 + 4x^2} dx so I let  x = \frac{1}{2}sinh(u)
    therefore  dx = \frac{1}{2}cosh(u)du


    subbed it in:  \frac{1}{2}\int cosh^2(u)du

    used identity to re-arrange:  \frac{1}{4} \int cosh(2u)du + \frac{1}{4}\int du

    evaluated: 2u) + \frac{1}{4}u + C " alt=" \frac{1}{8}sinh(2u) + \frac{1}{4}u + C " /> Mr F added the red 2.

    re arranged the substitution I made earlier for  u = arcsinh(2x)
    subbed u back in yielded  \frac{1}{8}sinh(arcsinh(2x)) + \frac{1}{4}arcsinh(2x) + C

    which cancelled to give  \frac{1}{4}x + \frac{1}{4} arcsinh(2x) + C

    however the answer I'm given is:  \frac{1}{4}(arcsinh(2x) + 2x\sqrt{1 + 4x^2}) + C

    Hopefully someone can spot what I've done wrong. Thanks!
    You should ask new question in a new thread.

    I think you've made the same mistake (missing the 2) as you did here: http://www.mathhelpforum.com/math-he...integrals.html

    Use the same approach.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    You should ask new question in a new thread.

    [...]
    Will do, was trying not to flood the forum with my threads though
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