Hi danagasta,
If my last instructions did not work then you need to reset your RAM. You are missing the negative sign and, I have no idea where you got the absolute value inside the brackets.
Yes, the absolute value are used in books and, on integral tables but, I can asure you that the Ti 89 should not have produce them in this answer.
I'll stand behind that In( -cos(x) / (sin(x)-1)) is the correct answer. You can go ahead and simplify after the calculation to get the answer you want but, that is the good answer that Ti 89 is going to give you for the integral of sec(x).
Errors that do not match the suspected, I am pretty sure that Texas Instruments would have you clear the RAM. I do not like resetting the RAM but, I could not find the cause to why you got the answer you did. And so far no other expert has produced a solution to convince me otherwise.
I was not able to come up with any of the 3 answers MiB8888 did on my Ti 89. I am not saying that they are not good answers, they are book and integral tables answers. You are not going to get them on the Ti 89.
See cap image below
Actually, it should be any of:
All three are equally correct, the TI89's solution is just a different way to write the formula than you expected.
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Are you sure there's not a - sign in front of cos(x) such as -cos(x). like in ln[ |-cos(x) / (sin(x)-1)| ] which is the correct answer.
I've tried everything to figure out what could be wrong to give you that answer. I'll keep tring and, let you know.
Where you getting the absolute from? That may be the problem. I'll see if I can assimulate.
I know you probably can't wait around for me to hit the right combination to give you a exact solution. It may have something to do with folder your working in or, not updated os. Make sure you start your problems with the NewProb command entered at the Home screen. Try working out of the main folder. Somethings corrupt.. I'll keep trying too.
The Ti89 microprocessor calculations are inversed g functions unless certain questions are answered first
Once you have simplified to ln[ |-cos(x) / (sin(x)-1)| ] , don't forget get the + C as in,ln[ |-cos(x) / (sin(x)-1)| ] + C
Or simplified to ln[ |(sin(x)-1)/ cos(x)| ] , don't forget get the + C as in, ln[ | (sin(x)-1)/ (cos(x)| ] + C . This may be the answer the instructor is looking for
Charles your work is correct, I just added a few things to it. g doesn't not mean guassian or something simple you might think
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