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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


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    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

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


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    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

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    [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].
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