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Math Help - integrals of secx and 1/cosx

  1. #1
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    integrals of secx and 1/cosx

    why are the integrals of secx and 1/cosx not the same??
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  2. #2
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    They are the same. Because sec(x)=1/cos(x)

    What difference do you see?. Please explain so I may know what you're getting at.
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  3. #3
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    the integral of secx=


    the integral of 1/cosx=

    ??
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  4. #4
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    They are equivalent, just in a different form.

    tan(\frac{x}{2}+\frac{\pi}{4})=sec(x)+tan(x)

    Can you prove it?.
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  5. #5
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    Quote Originally Posted by hmmmm View Post
    the integral of secx=


    the integral of 1/cosx=

    ??
    And what if \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\sec(ax)+\tan(ax) ?

    \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\frac{\tan \left(\frac{ax}{2}\right)+\tan \left(\frac \pi 4\right)}{1-\tan \left(\frac{ax}{2}\right)\tan \left(\frac \pi 4\right)}
    We know that \tan \frac \pi 4=1. hence :

    \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\frac{1+\tan \left(\frac{ax}{2}\right)}{1-\tan \left(\frac{ax}{2}\right)}

    Multiply by \frac{1+\tan \left(\frac{ax}{2}\right)}{1+\tan \left(\frac{ax}{2}\right)} :

    =\frac{\left(1+\tan \left(\frac{ax}{2}\right)\right)^2}{1-\tan^2 \left(\frac{ax}{2}\right)}

    Now substitute \tan^2t=\frac{1}{\cos^2t}-1

    Then multiply both numerator and denominator by cosē and use this formula : 2 \cos^2t-1=\cos(2t)

    This should help you go through the problem
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  6. #6
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    thanks

    thanks ive got it now
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