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Insights Solve Integrals Involving Tangent and Secant with This One Trick - Comments

  1. Mar 6, 2017 #1

    stevendaryl

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  2. jcsd
  3. Mar 6, 2017 #2
    Fascinating!

    The most commonly given proof in high school for integrating sec x is multiplying the numerator and denominator by sec x + tan x. To me it seemed like a highly unintuitive step. Integrating sec x by using partial fractions - a longer but more intuitive proof, originally discovered by Barrow is the one I prefer.

    This is pretty cool too. Thanks for the insight!
     
  4. Mar 6, 2017 #3

    vanhees71

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    I'd like to give this Insight 5 stars, but it just doesn't let me, because I'm not logged in for some reason when clicking at this specific Insight article. Anyway, it's a great trick, I hope to remember when I need it.
     
  5. Mar 7, 2017 #4

    vanhees71

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    Finally I could give you your 5 stars! :-).

    An alternative way to integrate ##\sec x## is to use the standard subsitution
    $$u=\tan \left (\frac{x}{2} \right), \quad \mathrm{d} x=\mathrm{d} u \frac{1}{1+u^2}.$$
    The integrand is
    $$\sec x= \frac{1}{\cos x}=\frac{1}{2 \cos^2 (x/2)-1}=\frac{1-u^2}{1+u^2}.$$
    Thus you get
    $$\int \mathrm{d} x \sec x= \int \mathrm{d} u \frac{2}{1-u^2} = \int \mathrm{d} u \left (\frac{1}{u+1}-\frac{1}{u-1} \right) = \ln \left |\frac{u+1}{u-1} \right|+C$$
    or finaly resubstituting
    $$\int \mathrm{d} x \sec x=\ln \left |\frac{\tan(x/2)+1}{\tan(x/2)-1} \right|+C.$$
     
  6. Mar 7, 2017 #5
    @vanhees71 Awesome, I hadn't seen this one.

    I still can't give five stars though.:sorry:
     
  7. Mar 7, 2017 #6
    Very cool Daryl, thanks!
     
  8. Mar 8, 2017 #7

    haushofer

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    Thanks, useful for my calculus teaching! :)
     
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