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

Anyone know how to integrate (Tanx)^2(Secx) thanks

  1. Dec 5, 2008 #1
    (Tanx)^2(Secx)
     
  2. jcsd
  3. Dec 5, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Not necessarily the simplest: tan(x)= sin(x)/cos(x) and sec(x)= 1/cos(x) so tan^2(x) sec(x)= sin^2(x)/cos^3(x). That involves cosine to an odd power so we can "factor out" one to use with a substitution. Specifically, multiply both numerator and denominator by cos(x) to get sin^2(x)cos(x)/cos^4(x). cos^4(x)= (cos^2(x))^2=(1- sin^2(x))^2 so let u= sin(x). Then du= cos(x) dx so the integral becomes
    [tex]\int \frac{u^2}{(1-u^2)^2} du[/tex]
    That integral can be done by partial fractions.
     
  4. Dec 5, 2008 #3
    Here's another substitution you could make, inspired by stereographic projection, and it always works when you have a rational function of trigonometric functions of x.

    Let [tex]u = \tan(x/2)[/tex]. From that, using various trigonometric identities, obtain the following:
    [tex]\sin x = \frac{2u}{1+u^2}; \cos x = \frac{1-u^2}{1+u^2}; dx = \frac{2 \,du}{1+u^2}[/tex]
    and in particular
    [tex]\tan x = \frac{2u}{1-u^2}; \sec x = \frac{1+u^2}{1-u^2}.[/tex]
    Then your integral becomes
    [tex]\int \tan^2 x \sec x \,dx = \int \frac{8u^2}{(1-u^2)^3} \,du[/tex]
    which can be done by partial fractions. Admittedly, you've ended up with a higher degree denominator than what HallsofIvy gets, but again it always reduces a rational function of trig functions to a rational function of u, which can be done by partial fractions (or if you're lucky, substitution).
     
    Last edited: Dec 5, 2008
  5. Dec 7, 2008 #4

    Gib Z

    User Avatar
    Homework Helper

    Expanding using Pythagorean Identities yields

    [tex]\int \tan^2 x \sec x dx = \int \sec^3 x dx - \int \sec x dx[/tex]

    The second integral, upon multiplying both the numerator and denominator by (sec x + tan x), evaluates to log |sec x + tan x|+ C

    The first requires integration by parts. You should be able to see the integral you want on both sides of the equation now, so isolate it and solve.
     
  6. Dec 7, 2008 #5

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    My calculus book gives an algorithm that can be used to solve all such kinds of trigonometric integrals. Doesn't yours have a similar thing?
     
  7. Dec 7, 2008 #6
    You're probably thinking of the u = tan(x/2) thing I mentioned above. :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Anyone know how to integrate (Tanx)^2(Secx) thanks
Loading...