1. Mar 8, 2006

### misskitty

I'm working through a definite integration with a secant function as my antiderivative...the secant is 1/cos right?

~Kitty

2. Mar 8, 2006

### TD

Yes, and 1/sin(x) is cosec(x) = csc(x).

3. Mar 8, 2006

### misskitty

Oh I forgot to ask this...is it possible to take a secant function on a calculator? My Calc teacher informed me that the secant function and 1/cos (or whatever the correct inverse function is) is not the same when you do them on the calculator. Once you do the initial function like cos, sin, tan is it wrong to talk that answer and divide one by it? If it isn't the same thing why is it different?

~Kitty

4. Mar 8, 2006

### misskitty

So it is 1/sin x. Any tips on keeping them straight to avoid brain fades? I do have them memorized, but there are times when I can't remember.

~Kitty

5. Mar 8, 2006

### TD

Ok, to avoid confusion (hopefully):

Secant: sec(x) = 1/cos(x)
Cosecant: csc(x) = 1/sin(x)

On a calculator though, they often write $\sin ^{ - 1}$ and $\cos ^{ - 1}$ for the inverse functions, so arcsin and arccos. Remember that this is only a notation (which I wouldn't use) and that it's not the same as 1/sin(x) and 1/cos(x).

6. Mar 8, 2006

### misskitty

Right. What I do when I have 1/sin x questions is I do the trig first so the sin x then once I have that answer I divide one by the answer I get. Is that still wrong?

~Kitty

7. Mar 8, 2006

### TD

That's ok, as well as doing 1/sin(x) immediately, just as long as you don't use the $\sin^{-1}$ button, since that's inverse sine and not 1/sin.

8. Mar 8, 2006

### misskitty

Why did he tell me I was doing this wrong? I explained my process exactly as I just did to you.

~Kitty

9. Mar 8, 2006

### TD

No idea, unless I'm misunderstanding you.

10. Mar 8, 2006

### misskitty

No I don't think you are.

The only other thing I can't figure out is why I keep getting the incorrect answer because I'm following the process of the Fundamental theorem of calculus the way my book tells me too down to the letter. Is there some reason why this keeps happening?

~Kitty

11. Mar 8, 2006

### TD

Perhaps you could post an example.

12. Mar 8, 2006

### VietDao29

Okay, what do you get for 1 / (sin(2.5)) (in radians mode)?
If you get an approximate answer of 1.67, then you are doing it correctly, or do you get a big ERROR message?

13. Mar 8, 2006

### misskitty

This might look kinda messy but this is one of the problems I've been working on sec^2xdx in the intervale of -pi/6 < x < pi/6. Does that make sense?

~Kitty

14. Mar 8, 2006

### misskitty

~Kitty

15. Mar 8, 2006

### misskitty

When I do 1/(sin 2.5) I get 1.69 for answer. How do you know when an answer is within limits? Like my answer is .02 different than yours. Does that mean its wrong?

~Kitty

16. Mar 8, 2006

### Hootenanny

Staff Emeritus
Yes, it makes sense. If the range in given in radians, you should use radians in you answers.

17. Mar 8, 2006

### VietDao29

Hmmm, not sure what you mean... What does the problem ask? What's the question?
--------------
By the way, 1.69 may be okay. There may be some rounding problem there...

18. Mar 8, 2006

### Hootenanny

Staff Emeritus
The 0.02 difference in likely to be a rounding error, try not to round in calculations by doing the whole calc on you calculator.

19. Mar 8, 2006

### misskitty

The problem is evaluating the definite integral.

BTW- I think my calculator might round automatically causing the .02 difference.

~Kitty

20. Mar 8, 2006

### VietDao29

$$\int \sec ^ 2 x dx = \tan x + C$$
It's the antiderivative for sec2x dx, I think you can find it in the list of integrals. Look at it in your book. Now apply:
$$\mathop{\int} \limits_{a} ^ b f(x) dx = F(b) - F(a)$$, and arrive at the answer.
You may want to skim through your textbook once again to get the main concept before tackling problems...
Can you go from here? :)