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Math Help - 2 Intergration questions

  1. #1
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    2 Intergration questions

    Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

    1)  \int\sqrt{1 - cos(2x})sin(2x) dx

    2)  \int\tan^{-1}(3z) dz

    Thanks
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  2. #2
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    Quote Originally Posted by StupidIdiot View Post
    Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

    1)  \int\sqrt{1 - cos(2x})sin(2x) dx

    2)  \int\tan^{-1}(3z) dz

    Thanks
    for the first use a u sub

    u=1-\cos(2x) \to du=2\sin(2x)dx


    for the 2nd integrate by parts with u=\tan^{-1}(3z)
    and dv=dx

    This should get you started.

    Good luck.
    Last edited by TheEmptySet; May 19th 2008 at 11:46 AM. Reason: lost a 2
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  3. #3
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    Hello,

    Quote Originally Posted by StupidIdiot View Post
    Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

    1)  \int\sqrt{1 - cos(2x})sin(2x) dx
    Substitute : u=1-\cos 2x \implies du=2\sin 2x dx \implies dx=\frac{du}{2 \sin 2x}

    --> \int \sqrt{1-\cos 2x} \cdot \sin 2x dx=\int \sqrt{u} \cdot \sin 2x \cdot \frac{1}{2\sin 2x} du=\int \frac 12 \cdot \sqrt{u} du

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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by StupidIdiot View Post
    Looks like things are getting busy around here! Exam time I suppose (same for me in 2 days). Well if anyone gets a chance could they help me with these two:

    1)  \int\sqrt{1 - cos(2x})sin(2x) dx

    2)  \int\tan^{-1}(3z) dz

    Thanks
    for the second one use integration by parts

    u=arctan(3z)

    dv=dz

    so \frac{3dz}{1+9z^2}

    amd v=z

    So \int{arctan(3z)dz}=arctan(3z)z-\int\frac{3z}{1+9z^2}dx

    \int\frac{3z}{1+9z^2}dz=\frac{1}{6}\int\frac{18z}{  1+9z^2}dz=\frac{\ln|1+9z^2|}{6}+C

    so \int{arctan(3z)dz}=z\cdot{arctan(3z)}-\frac{\ln|1+9z^2|}{6}+C
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  5. #5
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    Thanks

    Letting dv=dz is a genius/simple step that I hadn't thought of.

    Thanks to all three of you! Maybe now I can achieve my dream result (40%) tomorrow
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