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Math Help - Trig Substitutions

  1. #1
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    Trig Substitutions

    Let
    <br />
x = 3\sin u<br />

    <br />
\int {\frac{{dx}}<br />
{{x^2 \sqrt {9 - x^2 } }}}  = \frac{1}<br />
{9}\int {\csc ^2 udu}  =  - \frac{1}<br />
{9}(\frac{{\sqrt {9 - x^2 } }}<br />
{x}) + C<br />

    I understand everything until the end. Specifically,
    how to get

    <br />
(\frac{{\sqrt {9 - x^2 } }}<br />
{x})<br />

    Any help would be greatly appreciated. Thanks.
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  2. #2
    o_O
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    \frac{1}{9}\int csc^{2}udu = -\frac{1}{9} \cot u + C

    x = 3\sin u \quad \Rightarrow \quad u = \sin^{-1}\left(\frac{x}{3}\right)

    \int \frac{dx}{x^{2}\sqrt{9-x^{2}}} = -\frac{1}{9} \cot\left(\sin^{-1} \left(\frac{x}{3}\right)\right) + C

    Note that: \cot \left(\sin^{-1} x\right) = \frac{\sqrt{1-x^{2}}}{x}

    Can you see how it fits ?
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  3. #3
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    Quote Originally Posted by kid funky fried View Post
    Let
    <br />
x = 3\sin u<br />

    <br />
\int {\frac{{dx}}<br />
{{x^2 \sqrt {9 - x^2 } }}} = \frac{1}<br />
{9}\int {\csc ^2 udu} = - \frac{1}<br />
{9}(\frac{{\sqrt {9 - x^2 } }}<br />
{x}) + C<br />

    I understand everything until the end. Specifically,
    how to get

    <br />
(\frac{{\sqrt {9 - x^2 } }}<br />
{x})<br />

    Any help would be greatly appreciated. Thanks.
    <br />
x = 3\sin u \mbox{ then } dx=2\cos(u)<br />

    \int {\frac{{dx}}{{x^2 \sqrt {9 - x^2 } }}}

    \int \frac{3\cos(u)}{9\sin^2(u)\sqrt{9-9\sin^2{u}}}du

    \frac{1}{9}\int \frac{\cos(u)}{\sin^2(u)\sqrt{1-\sin^2{u}}}du

    \frac{1}{9}\int \frac{\cos(u)}{\sin^2(u)\sqrt{\cos^2{u}}}du

    \frac{1}{9}\int \frac{1}{\sin^2(u)}du

    \frac{1}{9}\int \csc^{u}du=-\frac{1}{9}\cot{u}+c

    we need to take a side trip for a moment

     x = 3\sin u \iff \frac{x}{3}=\sin(u) \iff \left( \frac{x}{3}\right)=\sin^{2}u \iff \left( \frac{x}{3}\right)^2= 1-\cos^2(u)

    \cos^2(u)=1-\left( \frac{x}{3}\right)^2 \iff \cos(u)=\sqrt{1-\left( \frac{x}{3}\right)^2}=\frac{1}{3}\sqrt{9-x^2}

    =-\frac{1}{9}\cot{u}+c=-\frac{1}{9}\frac{\cos(u)}{\sin(u)}+c=-\frac{1}{9}\frac{\frac{1}{3}\sqrt{9-x^2}}{\frac{x}{3}}+c

    -\frac{1}{9}\frac{\sqrt{9-x^2}}{x}+c

    wow I am tired.
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  4. #4
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    How did you know-

    <br />
\cot (\sin ^{ - 1} x) = \frac{{\sqrt {1 - x^2 } }}<br />
{x}<br />


    BTW, I was supposed to use a reference triangle.
    Thanks for your reply.
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  5. #5
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    \cot \theta = \frac{\cos \theta}{\sin \theta}

    \theta = \sin^{-1} x

    \cot \left(\sin^{-1} x\right) = \frac{{\color{red}\cos \left(\sin^{-1} x\right)}}{{\color{blue}\sin \left(\sin^{-1} x\right)}} = \frac{{\color{red}\sqrt{1-x^{2}}}}{{\color{blue}x}}
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  6. #6
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    Got it, I understand now. Thanks so much!
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  7. #7
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    A reciprocal substitution also works. (I dunno why people don't use it. )
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    A reciprocal substitution also works. (I dunno why people don't use it. )
    What is a reciporcal substitution? I might know it by a different name...but if you could explain that would be great
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  9. #9
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    x=\frac1u.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    x=\frac1u.
    ok...where would you go from there?
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  11. #11
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    Go from there by yourself, you'll see.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Go from there by yourself, you'll see.
    \int\frac{1}{\frac{1}{u^2}\sqrt{9-(\frac{1}{u})^2}}dx=..........hm..........integrations by parts? u=u^2... dv=\frac{1}{\sqrt{9-(\frac{1}{u})^2}}... du=2u... v=arcsin\bigg(\frac{1}{3u}\bigg)?
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int\frac{1}{\frac{1}{u^2}\sqrt{9-(\frac{1}{u})^2}}dx=..........hm..........integrations by parts? u=u^2... dv=\frac{1}{\sqrt{9-(\frac{1}{u})^2}}... du=2u... v=arcsin\bigg(\frac{1}{3u}\bigg)?
    ...I forgot dx
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  14. #14
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    That's an interesting sub o.O

    x = \frac{1}{u}
    dx = - \frac{1}{u^{2}}du

    \int \frac{-\frac{1}{u^{2}}du}{\frac{1}{u^{2}} \sqrt{9 - \frac{1}{u^{2}}}}

    = - \int \frac{du}{\sqrt{9 - \frac{1}{u^{2}}}}

     = -\frac{1}{3} \int \frac{du}{\sqrt{1 - \left(\frac{1}{3u}\right)^{2}}} = -\frac{1}{3} \arcsin \left(\frac{1}{3u}\right) + C
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  15. #15
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    It's really helpful, but not always works. In this cases you can turn the integral into an easy one.
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