Brain Fade About Trig w/Differential

[misskitty]

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

~Kitty

 

[TD]

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

~Kitty

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

 

[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

 

[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

 

[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).

 

[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

 

[TD]

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

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.

 

[misskitty]

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

~Kitty

 

[TD]

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

~Kitty

No idea, unless I'm misunderstanding you.

 

[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

 

[TD]

Perhaps you could post an example.

 

[VietDao29]

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

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?

 

[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

 

[misskitty]

I was in radians yes.

~Kitty

 

[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

 

[Hootenanny]

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

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

 

[VietDao29]

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

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...

 

[Hootenanny]

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.

 

[misskitty]

The problem is evaluating the definite integral.

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

~Kitty

 

[VietDao29]

The problem is evaluating the definite integral.

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

~Kitty

$$\int \sec ^ 2 x dx = \tan x + C$$
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? :)

 

[misskitty]

That exactly what I did to solve the problem. What did you come up with for an answer? My answer is 1.158 but the book says its wrong.

~Kitty

 

[VietDao29]

That exactly what I did to solve the problem. What did you come up with for an answer? My answer is 1.158 but the book says its wrong.

~Kitty

I get about 1.1547, so your answer seems correct. The book may be wrong. However, you should watch out for rounding error.
By the way, what does the book say?

 

[misskitty]

Let me check.

~Kitty

 

[misskitty]

2( square root of 3)/ 3 I believe they are equivalent answers. I think they might have mislabled the answers again. This book is notorious for doing that.

~Kitty

 

[VietDao29]

2( square root of 3)/ 3 I believe they are equivalent answers. I think they might have mislabled the answers again. This book is notorious for doing that.

~Kitty

One big hint:
What's
$$\tan \left( \frac{\pi}{6} \right)$$?
So:
$$\mathop{\int} \limits_{-\frac{\pi}{6}} ^ {\frac{\pi}{6}} \sec ^ 2 x dx = \tan \left( \frac{\pi}{6} \right) - \tan \left( \frac{-\pi}{6} \right) = \tan \left( \frac{\pi}{6} \right) + \tan \left( \frac{\pi}{6} \right) = 2 \tan \left( \frac{\pi}{6} \right) = ?$$
Can you go from here? :)

 

[misskitty]

I get how to do it and that is the process that I used exactly. I just need to know if my answer is correct. That example does tell me I'm going through the correct process though. I was walking blind through the book.

~Kitty

 

[VietDao29]

I get how to do it and that is the process that I used exactly. I just need to know if my answer is correct. That example does tell me I'm going through the correct process though. I was walking blind through the book.

~Kitty

Arrgghh, you've misinterpreted my post.
What I am trying to say is that $$\tan \left( \frac{\pi}{6} \right)$$ has a special value that you should remember.
It's:
$$\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$$
Now let's see if you can come up with the same answer as the book. :)

 

[misskitty]

Oh sorry. If you rationalize that it turns out to be 2(square root of 3)/ 3...

~Kitty

 

[misskitty]

I think that matches this crazy book.

~Kitty

 

[Hootenanny]

Oh sorry. If you rationalize that it turns out to be 2(square root of 3)/ 3...

~Kitty

How have you rationalised?

 

[misskitty]

Multiply 1/(square root of 3) by the (square root of 3)/(square root of 3). Leaving you with the square root of 3 in the numerator and 3 in the denominator. Then multiply that by 2/1 (same thing as 2) to give you 2(square root of 3)/3.

How did you think I rationalized?

 

[Hootenanny]

Why have you multiplied by 2/1? That has changed the fraction. Once you have mulitplied by $\frac{\sqrt{3}}{\sqrt{3}}$ the fraction is now rational. When you mulitplied by $\frac{2}{1}$ you changed the fraction!

 

[VietDao29]

Why have you multiplied by 2/1? That has changed the fraction. Once you have mulitplied by $\frac{\sqrt{3}}{\sqrt{3}}$ the fraction is now rational. When you mulitplied by $\frac{2}{1}$ you changed the fraction!

?
What? What do you mean? I just don't get this!
$$\mathop{\int} \limits_{-\frac{\pi}{6}} ^ {\frac{\pi}{6}} \sec ^ 2 x dx = \tan \left( \frac{\pi}{6} \right) - \tan \left( - \frac{\pi}{6} \right) = 2 \tan \left( \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$$
What misskitty did is, needless to say, completely correct!

 

[Hootenanny]

I apologise, it was just in her post she said she mulitplied by two, I didn't understand where it came from. I only saw the $\frac{1}{\sqrt{3}}$, not the $\frac{2}{\sqrt{3}}$