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Math Help - Curvelenght, what am I doing wrong?

  1. #1
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    Curvelenght, what am I doing wrong?

    I'm trying to calculate the curvlenght for y=ln(cosx) , 0 \leq x \leq \frac{\pi}{6}. But I can not figure out what I'm doing wrong.

    Using sec(x) is not allowed (because we don't learn that here in Sweden).

    Using formula: ds = \int \sqrt{1 + (\frac{dy}{dx})^2} dx where \frac{dy}{dx} = \frac{-sinx}{cosx}.

    \int \sqrt{ 1 + (\frac{dy}{dx})^2 } dx = \int \sqrt{ 1 + (\frac{-sinx}{cosx})^2 } dx  = \int \sqrt{ 1 + (tanx)^2 } dx =

    // 1+tan^2(x) = \frac{1}{cos^2(x)} //

    =\int \sqrt{\frac{1}{cos^2(x)}} dx = \int \frac{\sqrt{1}}{\sqrt{cos^2(x)}} dx = \int \frac{1}{cosx} dx =

    // \frac{1}{cosx} = \frac{cosx}{cos^2(x)} //

    = \int \frac{cosx}{cos^2x} dx =

    // cos^2(x) = 1- sin^2(x)  //

    = \int \frac{cosx}{ 1 - sin^2(x) } dx =  \int \frac{1}{1 - sin^2(x)} * cosx dx

    // Using  u = sinx ,  du = cosx dx //

    =   \int \frac{1}{1 - u^2} du = \int \frac{1}{(1 - u)(1+u)} du =

    // \frac{1}{(1 - u)(1+u)} = \frac{A}{1-u} + \frac{B}{1+u} \Rightarrow A+Au+B-Bu = 1 \Leftrightarrow A+B=1 and A-B=0 \Leftrightarrow A=\frac{1}{2} , B=\frac{1}{2} //

    \int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}( ln|1+u| + ln|1-u| ) =

    // ln(A) + ln(B) = ln(A*B) , (1+u)(1-u)=1-u^2 //

     = \frac{1}{2}ln|1-u^2| =

    // changing back u=sinx //

     = \frac{1}{2}ln|1-sin^2(x)| = \frac{1}{2}ln|cos^2(x)|  = ln|cosx|

    ds=\int \sqrt{ 1 + (\frac{-sinx}{cosx})^2 } dx = ln|cosx|

    Setting 0 \leq x \leq \frac{\pi}{6}

    But when calculating ln|cos( \frac{ \pi }{6})|-ln|1| I get -0,14... The correct answer should be (according to the book) \frac{1}{2}ln(3) which is about 0,5
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  2. #2
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    Quote Originally Posted by kallekall View Post
    I'm trying to calculate the curvlenght for y=ln(cosx) , 0 \leq x \leq \frac{\pi}{6}. But I can not figure out what I'm doing wrong.
    Go to this website Wolfram|Alpha
    In the input window, type in this exact phrase: integrate sqrt[1+tan[x]^2]dx. (you can copy & paste)
    Click the equals bar at the end if the input window.
    Be sure you click ‘show steps’ to see the solution.

    Quote Originally Posted by kallekall View Post
    Using sec(x) is not allowed (because we don't learn that here in Sweden).
    What is the point in that practice?
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  3. #3
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    Here's your problem

    \int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}(\ln|1+u| + \ln|1-u| )

    It should be

    \int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}( \ln|1+u| - \ln|1-u| )
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  4. #4
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    Smile Thank you!

    Quote Originally Posted by Plato View Post
    Quote Originally Posted by kallekall View Post
    Using sec(x) is not allowed (because we don't learn that here in Sweden).
    What is the point in that practice?
    Don't exactly understand what you ask, but Wolfram|Alpha is using sec(x) when calculateing the integral, and we don't get to learn how to use sec in Sweden.

    Quote Originally Posted by chengbin View Post
    Here's your problem

    \int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}(\ln|1+u| + \ln|1-u| )

    It should be

    \int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}( \ln|1+u| - \ln|1-u| )
    Thanks a lot! It took me a little time to understand why:
    \frac{1}{2} \int  \frac{1}{1-u} du = \frac{-1}{2}(\ln|1-u| ) but I get it now. Thanks again!
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