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Math Help - Trig Integral

  1. #1
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    Trig Integral

    Here is my original problem:

    \displaystyle\int 2\tan^{-1}(\frac{z}{4})dz

    \displaystyle u=2tan^{-1}(\frac{z}{4})

    \displaystyle du=\frac{2*\frac{1}{4}}{1+(\frac{z}{4})^2}dz=\frac  {\frac{1}{2}}{1+\frac{1}{16}z^2}dz

    \displaystyle dv=dz and \displaystyle v=z


    \displaystyle\int 2\tan^{-1}(\frac{z}{4})dz=2\tan^{-1}(\frac{z}{4})-\int \frac{\frac{1}{2}z}{1+\frac{1}{16}z^2}dz=2\tan^{-1}(\frac{z}{4})-ln(1+\frac{1}{16}z^2)+c

    Here is what the book got:

    \displaystyle =2\tan^{-1}(\frac{z}{4})-ln(1+\frac{1}{4}z^2)+c

    I know I must have made algebra mistake somewhere, I just cant figure it out where I went wrong. Please help! Thanks.
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  2. #2
    MHF Contributor
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    First, you forgot to multiply my z in the first section. uv-int vdu.
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  3. #3
    MHF Contributor
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    Also, you should have made a second u sub in the vdu integral.

    \displaystyle 2 z*tan^{-1}\left(\frac{z}{4}\right)-2\int\frac{4}{z^2+16}dz

    \displaystyle u=z^2+16 \ du=2zdz\rightarrow \frac{1}{2}du=zdz

    \displaystyle 2 z*tan^{-1}\left(\frac{z}{4}\right)-\frac{8}{2}\int\frac{du}{u}\rightarrow ....
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