Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 6, 2017 #1


    User Avatar
    Staff Emeritus
    Science Advisor

  2. jcsd
  3. Mar 6, 2017 #2

    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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    Thanks, useful for my calculus teaching! :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?