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Math Help - u-sub integration help

  1. #1
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    u-sub integration help

    guys first timer here. cant quite figure this one out.



    ʃ dt/ t√(5tē-3)


    it looks to me to be a sec^-1u type of question but i cant seem to find the correct u value. any help from you smart peoples?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by lj212 View Post
    guys first timer here. cant quite figure this one out.



    ʃ dt/ t√(5tē-3)


    it looks to me to be a sec^-1u type of question but i cant seem to find the correct u value. any help from you smart peoples?
     <br />
5t^2 - 3 = 3 \left(\frac{5}{3}t^2 - 1 \right) = 3 \left( \left(\frac{\sqrt{5}}{\sqrt{3}} t\right)^2 - 1\right)<br />
and u = \frac{\sqrt{5}}{\sqrt{3}} t
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by danny arrigo View Post
     <br />
5t^2 - 3 = 3 \left(\frac{5}{3}t^2 - 1 \right) = 3 \left( \left(\frac{\sqrt{5}}{\sqrt{3}} t\right)^2 - 1\right)<br />
and u = \frac{\sqrt{5}}{\sqrt{3}} t
    Since t=\frac{\sqrt{3}}{\sqrt{5}}u and dt = \frac{\sqrt{3}}{\sqrt{5}}\,du, you have: \int\frac{dt}{t\sqrt{5t^2-3}} = \int\frac{\frac{\sqrt{3}}{\sqrt{5}}\,du}{\frac{\sq  rt{3}}{\sqrt{5}}u\sqrt{3(u^2-1)}} = \frac{1}{\sqrt{3}}\int\frac{du}{u\sqrt{u^2-1}}

    Spoiler:
    =\frac{1}{\sqrt{3}}\sec^{-1}(u) = \boxed{\frac{1}{\sqrt{3}}\sec^{-1}\left(\sqrt{\frac{5}{3}}\cdot t\right)}
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