1. ## 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.

2. ## almost right

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.

3. 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)?

4. One integral is given by: $\displaystyle \int \cot^3 x \ dx \ = \ - \frac{1}{2}\cot^2 x - \ln |\sin x | + C_1$

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
• $\displaystyle \cot^2 x = \csc^2 x - 1$
• $\displaystyle \sin x = \left(\csc x\right)^{-1}$

For example, let's use the first identity:
\displaystyle \begin{aligned}- \frac{1}{2}\cot^2 x - \ln |\sin x | + C_1 & = -\frac{1}{2} \left(\csc^2 x - 1\right) - \ln |\sin x| + C_1 \\ &= -\frac{1}{2}\csc^2 x + \frac{1}{2} - \ln |\sin x | + C_1 \\ &= -\frac{1}{2}\csc^2 x - \ln |\sin x | + C_2 \qquad \text{We let: } C_2 = \frac{1}{2} + C_1 \end{aligned}

This is option (1). So (1) and (3) are antiderivatives of $\displaystyle \cot^3 x$.

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

_____________

Or you can simply differentiate all 4 to get $\displaystyle \cot^3 x$

5. 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???????

6. $\displaystyle \frac{d}{dx} \ln |u| = \frac{1}{u} \cdot u'$

So: $\displaystyle \frac{d}{dx} \ln | \csc x | = \frac{1}{\csc x} \cdot (\csc x)' = \frac{1}{\csc x} \cdot \left(-\csc x \cot x\right) = \cdots$

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

8. 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.

9. Yes you're right.

Recall the definition: $\displaystyle |a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0\end{cases}$

So: $\displaystyle |\ln x| = \begin{cases} \ln x & \text{if } \ln x \geq 0 \ \Leftrightarrow \ x \geq 1\\ -\ln x & \text{if } \ln x < 0 \ \Leftrightarrow \ x < 1 \end{cases}$

10. 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?

11. Yes. Remember, $\displaystyle f'(x)$ is essentially a limit. The limit from the RHS is not equal to the LHS so it does not exist at x = 1.