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Thread: secxtanx=1?

  1. #1
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    secxtanx=1?

    Can I use identities to make sec(x)tan(x)=1?



    Moderator edit: This question is not relevant to the Calculus subforum. However, the relevance of this thread to the Calculus subforum only becomes clear with post #3.
    Last edited by mr fantastic; Nov 16th 2010 at 03:03 PM.
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  2. #2
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    Quote Originally Posted by dbakeg00 View Post
    Can I use identities to make sec(x)tan(x)=1?
    What do you mean by 'make'?

    Do you mean solve $\displaystyle \sec(x)\tan(x)=1$ for $\displaystyle x~?$
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  3. #3
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    This is the wrong forum for this but my original problem is $\displaystyle \displaystyle\int\frac{secxtanx}{9+4sec^2x}dx$

    So I used the integral identity $\displaystyle \displaystyle\int\frac{1}{a^2+x^2}dx=\frac{1}{a}ta n^{-1}\frac{x}{a}+c$

    Then I rewrote the problem $\displaystyle \displaystyle\int\ secxtanx*\frac{1}{3^3+2^2sec^2x}dx$

    $\displaystyle \displaystyle\int secxtanx*\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$

    So, if i could find a way to transform that secxtanx, then I would be home free...is that clear?
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  4. #4
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    You do know that $\displaystyle \dfrac{{d\sec (x)}}{{dx}} = \sec (x)\tan (x)~?$
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  5. #5
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    Quote Originally Posted by Plato View Post
    You do know that $\displaystyle \dfrac{{d\sec (x)}}{{dx}} = \sec (x)\tan (x)~?$
    I do..and maybe I'm missing something because I'm still new to integrals but the answer given in the book is:

    $\displaystyle \displaystyle\frac{1}{6}tan^{-1}(\frac{2secx}{3})+c$

    which is what I come up with when I integrate this part

    $\displaystyle \displaystyle\frac{1}{\frac{1}{4}[(\frac{3}{2})^2+sec^2x]}dx$

    so I had the secxtanx left over and didnt know what to do with it
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  6. #6
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    $\displaystyle \displaystyle \int {\frac{{\sec (x)\tan (x)}}
    {{9 + 4\sec ^2 (x)}}dx = \frac{1}
    {4}\int {\frac{{\sec (x)\tan (x)}}
    {{\left( {\frac{3}
    {2}} \right)^2 + \sec ^2 (x)}}dx = \frac{1}
    {4}\left( {\frac{2}
    {3}} \right)\arctan \left( {\frac{{2x}}
    {3}} \right)} }
    $
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  7. #7
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    So what happened to the sec(x)tan(x)? shouldn't it be sec(x)?
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  8. #8
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    Quote Originally Posted by dbakeg00 View Post
    So what happened to the sec(x)tan(x)? shouldn't it be sec(x)?
    Let $\displaystyle u=\sec(x)$ then $\displaystyle \displaystyle \[\frac{1}
    {4}\int {\frac{{\sec (x)\tan (x)dx}}{{\left( {\frac{3}
    {2}} \right)^2 + \sec ^2 (x)}}} = \frac{1}{4}\int {\frac{{du}}
    {{\left( {\frac{3}{2}} \right)^2 + u^2 }}} $
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  9. #9
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    there goes the light bulb! Thanks, I really appreciate it!
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