TITLE: Arclength of the curve $y= \ln( \sec x)$ $ 0 \le x \le \pi/4$
QUESTION [6 upvotes]: Arclength of the curve $y= \ln( \sec x)$ $ 0 \le x \le \pi/4$
I know that I have to find its derivative which is easy, it is $\tan x$
Then I put it into the arclength formula
$$\int \sqrt {1 - \tan^2 x}$$
From here I am not sure what to do, I put it in wolfram and it got something massive looking. I know I can't use u substitution and I am pretty certain I have to algebraicly manipulate this before I can continue but I do not know how.

REPLY [5 votes]: As others have noted, it should be 
$$\int_0^{\pi/4} \sqrt{1+\tan^2 x} \ \ dx$$
Recall $1+\tan^2 x = \sec^2 x$
$$\int_0^{\pi/4} \sqrt{\sec^2 x} \ \ dx$$
Since all trig functions are positive in the first quadrant, we can simply rewrite the integrand as
$$\int_0^{\pi/4} {\sec x} \ \ dx$$
Which is a (relatively) well known integral that evaluates to
$$\log (\tan(x) + \sec (x))$$
Now simply evaluate at the endpoints - you should get around $.8814$ assuming I didn't make a button-punching error.