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Math Help - 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; November 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 \sec(x)\tan(x)=1 for x~?
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  3. #3
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    This is the wrong forum for this but my original problem is \displaystyle\int\frac{secxtanx}{9+4sec^2x}dx

    So I used the integral identity \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\int\ secxtanx*\frac{1}{3^3+2^2sec^2x}dx

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

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

    \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 \int {\frac{{\sec (x)\tan (x)}}<br />
{{9 + 4\sec ^2 (x)}}dx = \frac{1}<br />
{4}\int {\frac{{\sec (x)\tan (x)}}<br />
{{\left( {\frac{3}<br />
{2}} \right)^2  + \sec ^2 (x)}}dx = \frac{1}<br />
{4}\left( {\frac{2}<br />
{3}} \right)\arctan \left( {\frac{{2x}}<br />
{3}} \right)} } <br />
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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 u=\sec(x) then \displaystyle \[\frac{1}<br />
{4}\int {\frac{{\sec (x)\tan (x)dx}}{{\left( {\frac{3}<br />
{2}} \right)^2  + \sec ^2 (x)}}}  = \frac{1}{4}\int {\frac{{du}}<br />
{{\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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