Thread: Trigonometric integrals

I have a simple question. I will show you four expressions for a certain trigonometric integrals that I THINK are all exactly same but I'm not positive.

The original integral is INTEG: (cot x)^3

After simplifying it, you get INTEG: (csc x)^2 cotx - cotx

Here are the four expressions/results of the integral that I believe are the exactly same -

1. (-1/2)csc^2x - ln (abs(sinx)) + C

2. (-1/2)csc^2x - ln (abs(cscx)) + C

3. (-1/2)cot^2x - ln (abs(sinx)) + C

4. (-1/2)cot^2x - ln (abs(cscx)) + C


They're all the same, right? The integral of cot x isn't just ln (abs(sinx)) + C but also ln (abs(cscx)) + C right?

Right??? I'm not sure.

1 and 3 appear to be right, but I think 2 and 4 are slightly incorrect. The antiderivative of cot(x) is either ln(abs(sin(x))) + C or -ln(abs(csc(x))) + C. If you use the rules of logarithms you should be able to relate the two: if you consider that A*ln(x) = ln(x^A), you will find that ln(sin(x)) = ln(csc(x)^-1) = -ln(csc(x)). Thus, for 2 and 4, you should have (-1/2)csc^2x + ln (abs(cscx)) + C and (-1/2)cot^2x + ln (abs(cscx)) + C. When you integrate, you can always check by differentiating.

Ok, um............

Let's differentiate ln[abs(cscx)]

(d/dy)ln[abs(cscx)] = abs[(-cscx cotx)/(cscx)] (natural logarithm differentiation and since (d/dx)cscx = -cscx cotx)

abs[-cotx] canceled

cot x became positive because of the absolute value sign

This answer is the same that you get from differentiating ln[abs(sinx)] so maybe cot x has 2 integrals or what?

What am I missing....?


I know that ln(sinx) = -ln(cscx) but it looks like it doesn't matter because when I differentiated above I got cot x or maybe I'm forgetting some absolute value rules (which may cause the answer to be really -cotx)?

One integral is given by:

This is option 3. We can easily show that this is equivalent to the other options simply by using the fact that:

• Every antiderivative of f(x) differs by a constant
• 
• 

For example, let's use the first identity:


This is option (1). So (1) and (3) are antiderivatives of .

Try using the other identities and you will see the equivalences.

_____________

Or you can simply differentiate all 4 to get

Will you please show me the process of differentiating ln[abs(csc x)]? That's the point of this whole topic, NOT how to differentiate (cot x)^3. It should not equal cot x and I know I'm doing something wrong because whenever I try to differentiate it, I get cot x instead of - cot x. I showed my process in my above post...... anyone see anything wrong with it???????

So:

Aah, so when you differentiate with ln you lose the absolute value sign huh? Ok cool.

Ah, wait a minute. I understand why ln [abs(x)] loses its absolute value because it's irrational/undefined for it to have -x because it's not included in the graph of ln x.

So.... will you guys please show me the process of finding the derivative of [abs(lnx)]? I've a feeling that you have to do piecewise differentiation.

Yes you're right.

Recall the definition:

So:

Allright, thanks! Once you differentiate that, you get.............

f ' (x) = 1/x x > 1
= -1/x x < 1

None of the above expressions can exactly equal 1 because at x = 1 there's a cusp so it's not differentiable, right?

Yes. Remember, is essentially a limit. The limit from the RHS is not equal to the LHS so it does not exist at x = 1.