 In this video we provide the solution to question number seven for practice exam number two for math 1220, in which case we have to set up but not evaluate the integral for the length y equals the natural log of cosine of x from x equals zero to x equals pi thirds. I should also mention we're supposed to step in simplifying. Not, you know, again, we don't have to evaluate but we have to do some basic simplifications here. So when it comes to finding arc length, the basic formula members we're going to integrate ds for which ds can have a couple different formulas as our function the y coordinate is given in terms of the x coordinate. We would do ourselves a huge favor by taking the formula where we take the square root of one plus y prime squared dx. So that's the first thing we want to do here we want to calculate the derivative of y right here so dy dx. So notice with this one we have a chain of functions we have a natural log and a cosine the cosine is an inside. So the way you take the derivative of natural log you're going to put your inner function on the bottom. And then on the top you're going to put the derivative of the inner function which in this case is a negative sign of x for which notice here I can write that as a negative tangent of x. So that's going to make a very nice substitution right there. So plugging that in we're going to get the square root of one plus negative tangent quantity squared dx. I also want to consider the bounds but these were explicitly given to me zero to pi thirds like so. Now I mean this is the correct integral right here but again the instructions do ask us to simplify let's do some appropriate simplifications there. Now one things I want to note is of course that when you square a negative you would just get back to the positive version so one plus tangent squared. And so by the Pythagorean identity one plus tangent squared is actually the same thing as secant squared. So we end up with the square root of secant squared actually and that gives us our last simplification we would integrate from zero to pi thirds of the function secant. For which we can talk about the antiderivative secant that we wanted to but again this is just asked us to set up and simplify do not evaluate it. And so our simplified version without actually evaluating the antiderivative would be this integral from zero to pi thirds of secant x dx. Do remember for full credit here of course you do need to simplify you do need to differential and of course the balance should be correct as well.