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Math Help - [SOLVED] hm how does this integral work?

  1. #1
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    [SOLVED] hm how does this integral work?

    this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

    {integral} 2 cscx dx

    = 2 ln( cscx + cotx)

    can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.
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  2. #2
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    Quote Originally Posted by Legendsn3verdie View Post
    this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

    {integral} 2 cscx dx

    = 2 ln( cscx + cotx)

    can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.
    Yes, it is a clever use of 1. Multiply csc(x) by \frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}}

    And I believe the answer is -2 ln|cscx + cotx|
    Last edited by Chop Suey; October 9th 2008 at 12:43 PM.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    this one seems to be tricky? can some1 figure out what the book did between these two steps?>?

    {integral} 2 cscx dx

    = 2 ln( cscx + cotx)

    can anybody see what they did?> is it the clever use of 1? i m not exactly sure what went on there. any help appreciated.
    Yes, it is a clever use of 1:

    2\int\csc(x)\,dx=2\int\csc(x)\cdot\frac{\csc(x)+\c  ot(x)}{\csc(x)+\cot(x)}\,dx

    Simplify the integrand, and then apply the substitution u=\csc(x)+\cot(x)

    Take it from here

    --Chris
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  4. #4
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    No, it's nothing really.

    \int {\csc x}\ dx = -\ln |\csc x + \cot x|

    So:

     \int {2 \csc x}\ dx

     = 2 \int {\csc x}\   dx

     = 2 \cdot -\ln |\csc x + \cot x|

     = -2 \ln |\csc x + \cot x|.

    I hope that helps.

    me07.
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  5. #5
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    ok, i see what u did there, but how the heck did u know to use cscx+cotx for 1!!?!?
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Yes, it is a clever use of 1:

    2\int\csc(x)\,dx=2\int\csc(x)\cdot\frac{\csc(x)+\c  ot(x)}{\csc(x)+\cot(x)}\,dx

    Simplify the integrand, and then apply the substitution u=\csc(x)+\cot(x)

    Take it from here

    --Chris

    ok so u = cscx + cotx

    du = -cscx cotx - csc^2 x
    = -cscx cotx - (1 + cot^2 x)
    = -cscx cotx -1 - cot^2 x

    i m not getting the numerator for that substitution..



    edit nvm i had it right just went to far, i got it nvm thread solved ty.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    ok so u = cscx + cotx

    du = -cscx cotx - csc^2 x
    = -cscx cotx - (1 + cot^2 x)
    = -cscx cotx -1 - cot^2 x

    i m not getting the numerator for that substitution..
    Before you integrate, you have \int\frac{{\color{red}\csc^2x+\csc x\cot x}}{\csc x+\cot x}\,dx

    You found du to be \,du=\left(-\csc^2 x-\csc x\cot x\right) \,dx=-\left({\color{red}\csc^2 x+\csc x\cot x}\right)\,dx

    What integral should you get once you make this substitution?

    --Chris
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  8. #8
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    Quote Originally Posted by Chris L T521 View Post
    Before you integrate, you have \int\frac{{\color{red}\csc^2x+\csc x\cot x}}{\csc x+\cot x}\,dx

    You found du to be \,du=\left(-\csc^2 x-\csc x\cot x\right) \,dx=-\left({\color{red}\csc^2 x+\csc x\cot x}\right)\,dx

    What integral should you get once you make this substitution?

    --Chris
    yah hm i had it just took it to far, ty.
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