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Math Help - [SOLVED] Evaluate the Integral

  1. #1
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    [SOLVED] Evaluate the Integral

    I really need help figuring out this integral:
    integral 5x/sqrt(5-x^4)
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  2. #2
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    Sorry about the format of the question, I don't really know how to get the symbols in
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by miniwheats View Post
    I really need help figuring out this integral:
    integral 5x/sqrt(5-x^4)
    you can use a trig substitution: x^2 = \sqrt{5} \sin \theta

    if you don't like the x squared thing, you can do a regular substitution first... u = x^2


    Quote Originally Posted by miniwheats View Post
    Sorry about the format of the question, I don't really know how to get the symbols in
    see here
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  4. #4
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    Thanks a lot for your help!
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by miniwheats View Post
    Thanks a lot for your help!
    you're welcome
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  6. #6
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    Sorry, I just have one more question.
    \int\frac{3e^{2t} dt}{1+e^{4t}}

    I substituted u= e^{2t}, du=2e^{2t}dt to get:

    = \frac{3}{2}\int\frac{du}{1+u^2}

    = \frac{3}{2}\cdot\frac{1}{u}\arctan(\frac{1}{u})

    = \frac{3}{2}[\frac{1}{e^{2t}}\arctan(\frac{1}{e^{2t}})]

    which isn't the right answer, where did I go wrong?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by miniwheats View Post
    Sorry, I just have one more question.
    \int\frac{3e^{2t} dt}{1+e^{4t}}

    I substituted u= e^{2t}, du=2e^{2t}dt to get:

    = \frac{3}{2}\int\frac{du}{1+u^2}

    = \frac{3}{2}\cdot {\color{red} \frac{1}{u}\arctan(\frac{1}{u})}

    = \frac{3}{2}[\frac{1}{e^{2t}}\arctan(\frac{1}{e^{2t}})]

    which isn't the right answer, where did I go wrong?
    what is in red is your mistake. that formula is wrong. heck, you did not need that formula (the one you were thinking about).

    \int \frac {dx}{1 + x^2} = \arctan x + C
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  8. #8
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    Ack, my mistake. I made that question more complicated than it had to be by forgetting that \int\frac{dx}{1+x^2} = arctan(x) + C -_-

    Thanks again for your help!
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