# How do you find the arc length of the curve #y=lncosx# over the interval [0, pi/3]?

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Arc length is given by:

Simplify:

Integrate directly:

Insert the limits of integration:

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To find the arc length of the curve (y = \ln(\cos x)) over the interval ([0, \frac{\pi}{3}]), you use the arc length formula for a function (y = f(x)) over an interval ([a, b]):

[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

First, find (\frac{dy}{dx}):

[ \frac{dy}{dx} = \frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x} = -\tan x ]

Now, square it:

[ \left(\frac{dy}{dx}\right)^2 = \tan^2 x ]

Now, substitute into the arc length formula and integrate over the given interval:

[ L = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2 x} , dx ]

[ = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2 x} , dx ]

[ = \int_{0}^{\frac{\pi}{3}} \sqrt{\sec^2 x} , dx ]

[ = \int_{0}^{\frac{\pi}{3}} \sec x , dx ]

[ = \ln|\sec x + \tan x| \bigg|_{0}^{\frac{\pi}{3}} ]

[ = \ln|\sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3})| - \ln|\sec(0) + \tan(0)| ]

[ = \ln|\sqrt{3} + \frac{\sqrt{3}}{3}| - \ln|1 + 0| ]

[ = \ln\left(\frac{4\sqrt{3}}{3}\right) ]

So, the arc length of the curve (y = \ln(\cos x)) over the interval ([0, \frac{\pi}{3}]) is (\ln\left(\frac{4\sqrt{3}}{3}\right)).

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