`y = ln(cosx) , [0,pi/3]` Find the arc length of the graph of the function over the indicated interval.

The arc length of a function of x, f(x), over an interval is determined by the formula below:

`L=int_a^bsqrt(1+((dy)/(dx))^2)dx`

So using the function given, let us first find `(dy)/(dx):`

`d/(dx)(ln(cos(x)))=(1/(cos(x)))*(-sin(x))=-(sin(x))/(cos(x))=-tan(x)`

We can now substitute this into our formula above:

`L=int_a^bsqrt(1+((dy)/(dx))^2)dx=L=int_0^(pi/3)sqrt(1+(-tan(x))^2)dx`

Which can then be simplified to:

`L=int_0^(pi/3)sqrt(1+tan^2(x))dx=int_0^(pi/3)sqrt(sec^2(x))dx=int_0^(pi/3)sec(x)dx`

Then you find the definite integral as you normally would.  (Using the method shown on the link below, you can find the integral of sec(x).)

`L=int_0^(pi/3)sec(x)dx=ln|sec(x)+tan(x)|_0^(pi/3)`

`L=ln(sec(pi/3)+tan(pi/3))-ln(sec(0)+tan(0))=ln(2+sqrt(3))-ln(1+0)`

`L=ln(2+sqrt(3))-ln(1)=ln(2+sqrt(3))~~1.32`

So...

The arc length of a function of x, f(x), over an interval is determined by the formula below:

`L=int_a^bsqrt(1+((dy)/(dx))^2)dx`

So using the function given, let us first find `(dy)/(dx):`

`d/(dx)(ln(cos(x)))=(1/(cos(x)))*(-sin(x))=-(sin(x))/(cos(x))=-tan(x)`

We can now substitute this into our formula above:

`L=int_a^bsqrt(1+((dy)/(dx))^2)dx=L=int_0^(pi/3)sqrt(1+(-tan(x))^2)dx`

Which can then be simplified to:

`L=int_0^(pi/3)sqrt(1+tan^2(x))dx=int_0^(pi/3)sqrt(sec^2(x))dx=int_0^(pi/3)sec(x)dx`

Then you find the definite integral as you normally would.  (Using the method shown on the link below, you can find the integral of sec(x).)

`L=int_0^(pi/3)sec(x)dx=ln|sec(x)+tan(x)|_0^(pi/3)`

`L=ln(sec(pi/3)+tan(pi/3))-ln(sec(0)+tan(0))=ln(2+sqrt(3))-ln(1+0)`

`L=ln(2+sqrt(3))-ln(1)=ln(2+sqrt(3))~~1.32`

So the exact value of the arc length of the graph of the function over the given interval is `ln(2+sqrt(3))`

which is approximately 1.32.