1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculators TI-89 integral

  1. Mar 9, 2008 #1
    [SOLVED] TI-89 integral

    1. The problem statement, all variables and given/known data
    My TI-89 says that
    [tex]\int \frac{1}{\cos x} dx = \ln \left(\frac{-\cos(x)}{\sin(x)-1}\right)[/tex]
    which is just wrong isn't it!?!?!
    See http://en.wikipedia.org/wiki/Lists_of_integrals

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2
    don't trust wiki

    mine is giving me the same answer, but in absolute values... maybe your in a different format? Also, wikipedia can be edited by anyone so you probably shouldn't trust it for most things
  4. Mar 9, 2008 #3
    OK. Here is a better source https://www.math.lsu.edu/~adkins/m2065/IntegralTable.pdf. Even with absolute values the result is still clearly wrong as you can see from number 8 at the link. Wikipedia had the same result therefore I trust Wikipedia more than Texas Instruments :)

    EDIT: maybe it is not wrong; but then how do you manipulate the TI-89 answer to get the integral table answer?
  5. Mar 9, 2008 #4


    User Avatar
    Homework Helper

    It is correct as the expression in the brackets is equivalent to tanx+secx. But I don't have a TI-89 so I can't really help you. But why do you want it to show it in the way the integral table has it?
  6. Mar 9, 2008 #5
    well... i know that cos(x)/sin(x)= cot(x) and cot(x)-1= (cos(x)-sin(x))/sin(x)

    so.. i don't know they probably go together somehow
  7. Mar 10, 2008 #6

    Gib Z

    User Avatar
    Homework Helper

    Indefinite integrals of a function are unique only up to a constant.
  8. Mar 10, 2008 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    Well, I, not having a TI89 calculator at hand, would have to do the integral by hand (Poor me!).

    [tex]\int \frac{1}{cos x}dx= \int \frac{cos x}{cos^2 x} dx= \int \frac{cos x}{1- sin^2x}dx[/tex]
    Letting u= sin(x), du= dx, the integral becomes
    [tex]\int \frac{du}{1- u^2}= \frac{1}{2}\int \frac{du}{1-u}+ \frac{1}{2}\int \frac{du}{1+u}[/tex]
    by partial fractions. Integrating each of those, I get
    [tex]\frac{1}{2}ln(1-u)}+ \frac{1}{2}ln(1+u)}= ln\left(\sqrt{\frac{1+u}{1- u}}\right)+ C[/tex]
    [tex]= ln\left(\sqrt{\frac{1+ sin(x)}{1- sin(x)}}\right)+ C[/tex]
    Multiplying both numerator and denominator of the fraction by 1- sin(x),
    [tex]= ln\left(\sqrt\frac{1-sin^2(x)}{1+sin(x)^2}}\right)+ C= ln\left(\sqrt{\frac{cos^2(x)}{(1- sin2(x))^2}}\right)+ C[/itex]
    [tex]= ln\left(\frac{cos(x)}{1- sin(x)}\right)+ C[/tex]

    Yes, it looks like the TI89 is right!

    As far as the Wikipedia result is concerned, I suspect a trig identity would give the same thing but I will let someone else show that.
  9. Mar 10, 2008 #8

    Gib Z

    User Avatar
    Homework Helper

    [tex]\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{1+ \sin x}{\cos x}[/tex]

    [tex] \frac{\cos x}{1-\sin x} = \frac{\cos x (1+ \sin x)}{1-\sin^2 x} = \frac{\cos x(1+ \sin x)}{\cos^2 x} = \frac{1+\sin x}{\cos x}[/tex].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: TI-89 integral
  1. TI 89 Useful or not? (Replies: 0)