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Math Help - trig substitution

  1. #1
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    trig substitution

    Good afternoon,

    I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

     <br />
{\int{\frac{\sqrt{4-x^2}}{x^2}}dx<br />


    let x=2sin\theta
    let dx=2cos\theta d\theta


    I arrive at


    <br />
{\int{\frac{\sqrt{4-x^2}}{x^2}}<br />
=\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C
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  2. #2
    A Plied Mathematician
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    Hmm. I don't get your result. Could you please post a few more steps?
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  3. #3
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    Sure,

    Using the substition...

    <br />
x=2sin\theta<br />
    <br />
dx=2cos\theta d\theta<br />

    I get...

    <br />
\sqrt{4-x^2} = \sqrt{4-4sin^2\theta} = 2\sqrt{1-sin^2\theta} = 2\sqrt{cos^2\theta} = 2cos\theta<br />

    So...

     <br />
\int{\frac{2cos\theta}{4cos^2\theta}d\theta}<br />

    Wait... Should that have been...

    \int{\frac{d\theta}{4cos^2\theta}}

    This must have been where I went wrong. What should this have been, and why?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    [QUOTE=MechEng;530685]Good afternoon,

    I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

     <br />
{\int{\frac{\sqrt{4-x^2}}{x^2}}dx<br />


    let x=2sin\theta
    let dx=2cos\theta d\theta


    The substitution is good!!!

    ...but the final answer is wrong!

    (Take the derivative of what you got, is it the integrand?)
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  5. #5
    A Plied Mathematician
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    Double-check your denominator.
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  6. #6
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    Quote Originally Posted by MechEng View Post
    ...
     <br />
\int{\frac{2cos\theta}{4cos^2\theta}d\theta}<br />
    ...
    Your numerator is indeed 2\cos \theta but your denominator is x^2 = (2\sin \theta)^2, and don't forget to replace dx with 2\cos\theta d\theta.
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  7. #7
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    Totally not right.

    Quote Originally Posted by MechEng View Post
    <br />
{\int{\frac{\sqrt{4-x^2}}{x^2}}<br />
=\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C
    [LaTeX ERROR: Convert failed]
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    no,

    it's:

    \int cot^2\theta d\theta = -\theta- cot\theta +C
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  9. #9
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    Quote Originally Posted by Also sprach Zarathustra View Post
    No
    What? The OP has not yet responded. I don't know what on Earth you are referring to.

    it's:

    \int cot^2\theta d\theta = -\theta + cot\theta +C
    The plus should be minus.

    EDIT. I see you edited your post.
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    You are blind or what??? Where are you see minus??? hahaha...
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  11. #11
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    Good afternoon All,

    I am sitting at work without my notes today, trying to pickup where I left off on this one.

    Using the "correct" substitution I get the following:

     <br />
\int{\frac{2cos\theta}{(2sin\theta)^2}(2cos\theta)  d\theta<br />

    Can this be reduced to:

     <br />
\int{\frac{(2cos\theta)^2}{(2sin\theta)^2}d\theta<br />

    And then:

     <br />
\int{cot^2\theta}d\theta<br />

    And then:

     <br />
-\theta-cot^2\theta+c<br />

    Since x=2sin\theta we have cot\theta=\frac{\sqrt{4-x^2}}{x} and \theta=sin^{-1}(\frac{x}{2})

    So:

    -sin^{-1}(\frac{x}{2})-(\frac{\sqrt{4-x^2}}{x})^2+C

    And:

     <br />
-sin^{-1}(\frac{x}{2})-\frac{\mid4-x^2\mid}{x^2}+C<br />


    Am I totally off target again?
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  12. #12
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    Quote Originally Posted by MechEng View Post
     <br />
\int{cot^2\theta}d\theta<br />

    And then:

     <br />
-\theta-cot^2\theta+c<br />
    Are you sure?
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  13. #13
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    Should that look more like:

    \int{cot^2\theta}d\theta = \ln{\mid sin^2\theta\mid+C

    I guess I'm confused by the cot^2\theta. Well, that and I am trying to do this without notes of any kind... between phone calls and e-mails.

    Thank you.
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  14. #14
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    note the identity ...

    \displaystyle \cot^2{t} = \csc^2{t} - 1

    sub in for \cot^2{t} in the integrand ...

    \displaystyle \int \csc^2{t} - 1 \, dt = -\cot{t} - t + C

    \displaystyle t = \arcsin\left(\frac{x}{2}\right)

    back substitute ...

    \displaystyle -\frac{\sqrt{4-x^2}}{x} - \arcsin\left(\frac{x}{2}\right) + C
    Last edited by skeeter; June 29th 2010 at 01:32 PM. Reason: fixed typo
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  15. #15
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    Quote Originally Posted by skeeter View Post

    sub in for \cos^2{t} in the integrand ...
    Where am I substituting this?

    Oh, should that have read:

    Quote Originally Posted by skeeter View Post

    sub in for \cot^2{t} in the integrand ...
    I'm getting really hung up on my trig identities.
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