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Math Help - integration by substitution

  1. #1
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    integration by substitution

    Hi there,

    I've been given the problem attached and its causing me lots of problems. When I perform the substitution I don't know how to "correct" du so that the new integral has a denominator that is equivalent to t-2.
    Please help!
    Attached Thumbnails Attached Thumbnails integration by substitution-photo-07-05-2013-10-54-20.jpg  
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: integration by substitution

    With the given substitution:

    u=\sqrt{t-1}\,\therefore\,du=\frac{1}{2u}\,dt\,\therefore\,d  t=2u\,du

    If we square the substitution:

    u^2=t-1\,\therefore\,t-2=u^2-1

    Can you proceed?
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  3. #3
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    Re: integration by substitution

    Quote Originally Posted by flashylightsmeow View Post
    Hi there,

    I've been given the problem attached and its causing me lots of problems. When I perform the substitution I don't know how to "correct" du so that the new integral has a denominator that is equivalent to t-2.
    Please help!
    \displaystyle \begin{align*} \int{ \frac{ \sqrt{ t - 1 } }{ t - 2 } \, dt } &= 2 \int{ \frac{ t - 1 }{ 2 \, \sqrt{ t - 1} \left( t - 2 \right) } \, dt }  \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = \sqrt{ t - 1} \implies du = \frac{1}{2 \, \sqrt{t - 1} } \, dt \end{align*} and the integral becomes

    \displaystyle \begin{align*} 2 \int{ \frac{ u^2 }{ u^2 - 1 } \, du } &= 2 \int{ 1 + \frac{1}{u^2 - 1} \, du} \end{align*}

    And now you need to apply Partial Fractions to continue.
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  4. #4
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    Re: integration by substitution

    AH YES! Thank you very much!!!! I was careless with my substitution!
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