TI-89 Integral No More Confusion

In summary, the conversation discusses the solution to an integral on a TI-89 calculator, which is believed to be incorrect by one person. Another person provides a link to a reliable source for integrals and explains that the result is correct, as it is equivalent to another expression. The conversation also includes a handwritten solution to the integral and a discussion on Wikipedia's result. It is determined that the TI-89's answer is correct and it can be manipulated to match the result on the integral table by using a trigonometric identity.
  • #1
ehrenfest
2,020
1
[SOLVED] TI-89 integral

Homework Statement


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

Homework Equations


The Attempt at a Solution

 
Last edited:
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  • #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
 
  • #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?
 
  • #4
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?
 
  • #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
 
  • #6
Indefinite integrals of a function are unique only up to a constant.
 
  • #7
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.
 
  • #8
[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].
 

Related to TI-89 Integral No More Confusion

1. What is the "TI-89 Integral No More Confusion" program?

The "TI-89 Integral No More Confusion" program is a software application designed for the TI-89 graphing calculator. It helps users solve complex integrals by providing step-by-step solutions and explanations.

2. How does the program work?

The program uses advanced algorithms and mathematical rules to break down the integral into smaller, more manageable steps. It then guides the user through each step, showing the necessary calculations and providing explanations along the way.

3. Is the program accurate?

Yes, the program is highly accurate. It uses proven mathematical methods and follows the same principles as solving integrals by hand. However, it is always recommended to double-check the results and use the program as a learning tool.

4. Can the program solve all types of integrals?

The program can solve a wide range of integrals, including definite and indefinite integrals, trigonometric integrals, and even improper integrals. However, there may be some rare cases where the program is unable to solve a specific type of integral.

5. Is the program user-friendly?

Yes, the program is designed to be user-friendly and easy to use. It has a simple and intuitive interface, and the step-by-step instructions make it accessible for users of all levels of mathematical knowledge.

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