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Math Help - integrating trig function

  1. #1
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    integrating trig function

    Can someone please explain to me how the

     \int secx = ln(sec x + tan x)

    am not even sure how to go about it, but  \int sec x = \int \frac{1}{cos x} ?


     \int (cosx)^{-1} ? And am not sure how to go from here, any help appreciated.
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  2. #2
    Member Goku's Avatar
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    Re: integrating trig function

    You could times sec(x) by \frac{sec(x)+tan(x)}{sec(x)+tan(x)}... and then solve from there...
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  3. #3
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    Re: integrating trig function

    Here is another approach.

    \int \frac{1}{\cos x} \;dx = \int \frac{\cos x}{\cos^2 x} \;dx = \int \frac{\cos x}{1 - \sin^2x}\; dx = \int \frac{du}{1-u^2}

    where we made the substitution u = \sin x. Now apply partial fractions and finish up.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: integrating trig function

    Another alternative could be to use the substitution x=\arcsin(t) \Rightarrow dx=\frac{dt}{\sqrt{1-t^2}} and using the fact that \cos[\arcsin(t)]=\sqrt{1-t^2} the integral becomes:
    \int \frac{dx}{\cos(x)}= \int \frac{\frac{dt}{\sqrt{1-t^2}}}{\sqrt{1-t^2}}=\int \frac{dt}{1-t^2}=\int \frac{dt}{(1-t)(1+t)} = \ldots
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