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  1. #1
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    antiderivate

    how do I find the antiderivate to:

    cot(x+4)

    and

    xarccot(4x) ??
    Last edited by weasley74; March 11th 2008 at 10:18 AM.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by weasley74 View Post
    how do I find the antiderivate to:

    cot(x+4)

    and

    xarctan(4x) ??
    cot(x+4)=\frac{cos(x+4)}{sin(x+4)}

    Set u=sin(x+4)
    du=cos(x+4)dx

    \int \frac{cos(x+4)}{sin(x+4)} dx = \int \frac{du}{u} \Rightarrow ln|u| + C

    Sub u=sin(x+4) back in.
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  3. #3
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    I'm sorry! it should be (cot(x+4))^2
    Last edited by ThePerfectHacker; March 16th 2008 at 07:23 PM.
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  4. #4
    Super Member wingless's Avatar
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    \int \cot^2(x+4)~dx

    \int \frac{\cos^2(x+4)}{\sin^2(x+4)}~dx

    Let u = x + 4, then du = dx.

    \int \frac{\cos^2 u}{\sin^2 u}~du

    \int \frac{1 - \sin^2 u}{\sin^2 u}~du

    \int \frac{1}{\sin^2 u}~du - \int 1~du

    \int \csc^2 u ~du - \int 1~du

    -\cot u - u + C

    -\cot(x+4) - x - 4 + C

    Or simply,

    -\cot(x+4) - x

    -[x+\cot(x+4)]
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by colby2152 View Post
    cot(x+4)=\frac{cos(x+4)}{sin(x+4)}

    Set u=sin(x+4)
    du=cos(x+4)dx

    \int \frac{cos(x+4)}{sin(x+4)} dx = \int \frac{du}{u} \Rightarrow ln|u| + C

    Sub u=sin(x+4) back in.
    Interesting note, subbing u=sin(x+4) back in gives:

    ln|sin(x+4)| + C

    However, quick integration rules says the integral is:

    -csc^2(x+4) + C

    Setting these equal to each other, since they are equivalent gives:

    ln|sin(x+4)| = -csc^2(x+4)
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