# Finding the are of the curve

• Mar 1st 2010, 04:25 PM
racewithferrari
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:
Attachment 15714
• Mar 1st 2010, 04:45 PM
skeeter
Quote:

Originally Posted by racewithferrari
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:
Attachment 15714

$\displaystyle \sqrt{sec^2{x}} = |\sec{x}|$

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

further, note that ...

$\displaystyle \int \sec{x} \, dx = \ln|\sec{x} + \tan{x}| + C$
• Mar 1st 2010, 05:48 PM
racewithferrari
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.
• Mar 2nd 2010, 06:46 AM
skeeter
you say this in your original post ...

Quote:

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

Quote:

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 ...

$\displaystyle \int_0^{\frac{\pi}{3}} \sec{x} \, dx$

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

$\displaystyle \left[\ln\left(\sec{\frac{\pi}{3}} + \tan{\frac{\pi}{3}}\right)\right] - \left[\ln\left(\sec{0} + \tan{0}\right)\right]$

$\displaystyle \left[\ln\left(2 + \sqrt{3}\right)\right] - \left[\ln\left(1+0\right)\right]$

$\displaystyle \ln(2 + \sqrt{3})$