Results 1 to 4 of 4

Math Help - Finding the are of the curve

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    92

    Finding the are of the curve

    Find length of the curve
    y=ln(cosx) {x,0,pi/2}

    I am stuck and I dont know how to integrate sqrt(sec^2).

    My Work:
    Finding the are of the curve-ques-10-2-.png
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by racewithferrari View Post
    Find length of the curve
    y=ln(cosx) {x,0,pi/2}

    I am stuck and I dont know how to integrate sqrt(sec^2).

    My Work:
    Click image for larger version. 

Name:	ques(10+2).PNG 
Views:	26 
Size:	459.3 KB 
ID:	15714
    \sqrt{sec^2{x}} = |\sec{x}|

    since the limits of integration are in quad I, you can drop the absolute value.

    further, note that ...

    \int \sec{x} \, dx = \ln|\sec{x} + \tan{x}| + C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    92
    I do not get it about dropping the absolute value.
    I know what the integral of sqrt(sec^2) is . But I know dont know how I will solve it comes to Exams.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    you say this in your original post ...

    I am stuck and I dont know how to integrate sqrt(sec^2).
    then you say this in a subsequent post ...

    I know what the integral of sqrt(sec^2) is. But I know dont know how I will solve it comes to Exams.
    pardon me if I do not understand your difficulty.

    here is the arclength calculation ...

     <br />
\int_0^{\frac{\pi}{3}} \sec{x} \, dx<br />

    \left[\ln(\sec{x} + \tan{x})\right]_0^{\frac{\pi}{3}}

     <br />
\left[\ln\left(\sec{\frac{\pi}{3}} + \tan{\frac{\pi}{3}}\right)\right] - \left[\ln\left(\sec{0} + \tan{0}\right)\right]<br />

     <br />
\left[\ln\left(2 + \sqrt{3}\right)\right] - \left[\ln\left(1+0\right)\right]<br />

     <br />
\ln(2 + \sqrt{3})<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding peaks on a curve
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 10th 2011, 06:23 AM
  2. Replies: 1
    Last Post: July 3rd 2010, 10:40 PM
  3. Finding a cosine curve from a sine curve?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 29th 2009, 03:40 AM
  4. Replies: 1
    Last Post: April 9th 2009, 09:02 AM
  5. Replies: 8
    Last Post: October 8th 2007, 04:29 PM

Search Tags


/mathhelpforum @mathhelpforum