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Math Help - Trigonometric integrals

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    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)?
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  4. #4
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    One integral is given by: \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
    • \cot^2 x = \csc^2 x - 1
    • \sin x = \left(\csc x\right)^{-1}


    For example, let's use the first identity:
    \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 \cot^3 x.

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

    _____________

    Or you can simply differentiate all 4 to get \cot^3 x
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  5. #5
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    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???????
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  6. #6
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    \frac{d}{dx} \ln |u| = \frac{1}{u} \cdot u'

    So: \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
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  7. #7
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    Aah, so when you differentiate with ln you lose the absolute value sign huh? Ok cool.
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  8. #8
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    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.
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  9. #9
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    Yes you're right.

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

    So: |\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}
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  10. #10
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    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?
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  11. #11
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    Yes. Remember, 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.
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