TI-89 Integral No More Confusion

[ehrenfest]

[SOLVED] TI-89 integral

Homework Statement

My TI-89 says that
$$\int \frac{1}{\cos x} dx = \ln \left(\frac{-\cos(x)}{\sin(x)-1}\right)$$
which is just wrong isn't it??!
See http://en.wikipedia.org/wiki/Lists_of_integrals

Homework Equations

The Attempt at a Solution

 

[feuerwasser]

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

 

[ehrenfest]

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?

 

[rock.freak667]

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?

 

[feuerwasser]

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

 

[Gib Z]

Indefinite integrals of a function are unique only up to a constant.

 

[HallsofIvy]

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

$$\int \frac{1}{cos x}dx= \int \frac{cos x}{cos^2 x} dx= \int \frac{cos x}{1- sin^2x}dx$$
Letting u= sin(x), du= dx, the integral becomes
$$\int \frac{du}{1- u^2}= \frac{1}{2}\int \frac{du}{1-u}+ \frac{1}{2}\int \frac{du}{1+u}$$
by partial fractions. Integrating each of those, I get
$$\frac{1}{2}ln(1-u)}+ \frac{1}{2}ln(1+u)}= ln\left(\sqrt{\frac{1+u}{1- u}}\right)+ C$$
$$= ln\left(\sqrt{\frac{1+ sin(x)}{1- sin(x)}}\right)+ C$$
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]<br /> $$= ln\left(\frac{cos(x)}{1- sin(x)}\right)+ C$$<br /> <br /> Yes, it looks like the TI89 is right!<br /> <br /> 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.[/tex]

 

[Gib Z]

$$\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{1+ \sin x}{\cos x}$$

$$\frac{\cos x}{1-\sin x} = \frac{\cos x (1+ \sin x)}{1-\sin^2 x} = \frac{\cos x(1+ \sin x)}{\cos^2 x} = \frac{1+\sin x}{\cos x}$$.