# Find the length of the curve: y = ln(sec x) from x = 0 to x = pi/4.

## Question:

Find the length of the curve: {eq}y = \ln(\sec x) {/eq} from {eq}x = 0 {/eq} to {eq}x = \frac{\pi}{4} {/eq}.

## Length of Curve:

The length of the curve can be found by using the formula {eq}L=\int_{a}^{b}\sqrt{1+y'^{2}}dx {/eq} where we will differentiate the curve and then plug-in the values in the formula and then plug in the upper and lower bounds.

## Answer and Explanation:

To solve the length of the curve we will proceed as

{eq}L=\displaystyle\int_{a}^{b}\sqrt{1+y'^{2}}dx {/eq}

where a,b is the interval

{eq}y=\displaystyle\ln sec x\\ \displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\sec x\tan x}{\sec x}\\ =\displaystyle\tan x {/eq}

The length of the curve will be

{eq}L=\displaystyle\int_{0}^{\frac{\pi}{4}}\sqrt{1+\tan^{2}x}dx\\ =\displaystyle\int_{0}^{\frac{\pi}{4}}\sec xdx\\ =\displaystyle\left \ln |\sec x+\tan x| \right ] {/eq}

Now plug-in the upper and lower bounds

={eq}\displaystyle\ln |1+\sqrt{2}| {/eq}