# Thread: Curvelenght, what am I doing wrong?

1. ## 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

2. Originally Posted by kallekall
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.

Originally Posted by kallekall
Using sec(x) is not allowed (because we don't learn that here in Sweden).
What is the point in that practice?

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

It should be

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

4. ## Thank you!

Originally Posted by Plato
Originally Posted by kallekall
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.

Originally Posted by chengbin
$\int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}(\ln|1+u| + \ln|1-u| )$
$\int \frac{1}{ 1 - u^2 } du = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u} du = \frac{1}{2}( \ln|1+u| - \ln|1-u| )$
$\frac{1}{2} \int \frac{1}{1-u} du = \frac{-1}{2}(\ln|1-u| )$ but I get it now. Thanks again!