Curvelenght, what am I doing wrong?

**kallekall** · July 23rd 2009, 06:53 AM

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

**Plato** · July 23rd 2009, 07:16 AM

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?

**chengbin** · July 23rd 2009, 07:38 AM

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

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

**kallekall** · July 23rd 2009, 08:49 AM

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

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!

Thread: Curvelenght, what am I doing wrong?

LinkBack

Thread Tools

Display

Curvelenght, what am I doing wrong?

Thank you!

Similar Math Help Forum Discussions

[SOLVED] am I wrong here or is my textbook wrong??

What am I doing wrong?

how can this be wrong?

My teacher says this linear equation is wrong, I think she is wrong.

Wrong answer key or wrong me ?

Search Tags