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Math Help - Indefinite Integrals

  1. #1
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    Indefinite Integrals

    Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

    Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.
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    Re: Indefinite Integrals

    Quote Originally Posted by nav92 View Post
    Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

    Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.
    To get dy/dx, use the chain rule (you will also need to use the quotient rule during this process).

    I suggest you review some examples from your class notes or textbook (integration by recognition is the name this 'technique' is often given).

    If you need more help, please show what you've done and say where you get stuck.
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    Re: Indefinite Integrals

    Alright, I know the how the chain rule works and here goes....

    y= ln u ; u=cotx

    dy/du= 1/u ; du/dx= -cosec^2 x

    but why dy/dx cotx = -cosec^2 x . I don't get it.
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    Re: Indefinite Integrals

    Quote Originally Posted by nav92 View Post
    Alright, I know the how the chain rule works and here goes....

    y= ln u ; u=cotx

    dy/du= 1/u ; du/dx= -cosec^2 x

    but why dy/dx cotx = -cosec^2 x . I don't get it.
    Trivial, either it is a standard derivative and so you should know it, or write:

    (d/dx)[cot(x)]=(d/dx)[cos(x)/sin(x)]=-1/sin^2(x)


    CB
    Last edited by CaptainBlack; June 22nd 2011 at 05:44 AM.
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    Re: Indefinite Integrals

    Quote Originally Posted by nav92 View Post
    Given that y= ln cot x, find dy/dx. Hence, find ∫(2-cosec 2x)dx.

    Okay basically, I'm stuck on how to differentiate y=ln cot x & then inserting it into the integration equation.
    You might like to consider using brackets around function arguments, and "^" to denote raising to a power.

    CB
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    Re: Indefinite Integrals

    Quote Originally Posted by CaptainBlack View Post
    cot(x)=cos(x)/sin(x)=(d/dx)[-1/sin^2(x)]
    Typo - rather :

    (d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]

    And, just in case a picture helps with this part...



    Balloon Calculus: standard integrals, derivatives and methods
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    Re: Indefinite Integrals

    Quote Originally Posted by tom@ballooncalculus View Post
    Typo - rather :

    (d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]
    fixed

    CB
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    Re: Indefinite Integrals

    Quote Originally Posted by tom@ballooncalculus View Post
    Typo - rather :

    (d/dx)cot(x)=(d/dx)[cos(x)/sin(x)]=[-1/sin^2(x)]

    And, just in case a picture helps with this part...



    Balloon Calculus: standard integrals, derivatives and methods
    Perfect. That diagram really sparked my understanding in the derivation of cosec^2 (x).

    Well now how to get dy/dx. I know it should be;
    dy/dx= (1/cot x)*( -cosec^2 (x) )
    = ??
    The answer suppose to be dy/dx= -2 cosec 2x but I don't know how to get it?
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    Re: Indefinite Integrals

    Quote Originally Posted by nav92 View Post
    dy/dx= (1/cot x)*( -cosec^2 (x) )
    = ??
    ... is correct. Then just say all of that in terms of sin and cos, and then apply the trig identity that gives 2 sin x cos x. (Know the one?)
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  10. #10
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    Re: Indefinite Integrals

    Quote Originally Posted by tom@ballooncalculus View Post
    ... is correct. Then just say all of that in terms of sin and cos, and then apply the trig identity that gives 2 sin x cos x. (Know the one?)
    Which trig identity are you referring to? I'm not sure.
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  11. #11
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    Re: Indefinite Integrals

    sin(2x) = 2 sinx cosx
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    Re: Indefinite Integrals

    Quote Originally Posted by tom@ballooncalculus View Post
    sin(2x) = 2 sinx cosx
    Okay, now I'm not sure where should I put the sin(2x)= 2sinxcosx to get 2-cosec(2x). Pls elaborate on this.

    dy/dx= (1/cot x)*( -cosec^2 (x) )
    = ??

    Thank you.
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  13. #13
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    Re: Indefinite Integrals

    Hi there,

    \frac{1}{\cot(x)}\ (-\csc^2(x))

    =\ \tan(x)\ \frac{-1}{\sin^2(x)}

    =\ \frac{- \sin(x)}{\cos(x) \sin^2(x)}

    =\ \frac{-1}{\sin(x) \cos(x)}

    =\ \frac{-1}{\frac{1}{2} \sin(2x)}

    =\ \frac{-2}{\sin(2x)}

    PS you have a typo in your latest (i.e. should be -2 not 2-, maybe that was the bug).
    Last edited by tom@ballooncalculus; June 23rd 2011 at 07:46 AM. Reason: ps
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  14. #14
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    Re: Indefinite Integrals

    Roger that, thank you.
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