Arclength of the curve $y= \ln( \sec x)$ $ 0 \le x \le \pi/4$ 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.
 A: You made a mistake. Arc length is given by the integral:
$$
\ell = \int_a^b \sqrt{1+\left(y'\right)^2}\,dx
$$
So if $y' = \tan x$, arc length is:
$$
\ell = \int_a^b \sqrt{1+\tan^2 x}\,dx = \int_a^b \sqrt{\sec^2 x}\,dx = \int_a^b |\sec x| \,dx
$$
For $a = 0$, $b = \frac{\pi}{4}$:
$$
\ell = \int_0^{\pi/4} \sec x \,dx
$$
The absolute value is gone as $\sec x$ is positive in $[0, \frac{\pi}{4}]$.
A: The arclength formula is 
$$\mathrm S_a^b(f) =\int_a^b \sqrt{1+f'(x)^2}dx$$ 
You have
$$f(x) = \log \sec x$$
This means 
$$f'(x) = \tan x$$
Then you need to find
$$\mathrm S =\int_0^{\pi/4} \sqrt{1+\tan^2 x}dx$$ 
Remember that
$$1+\tan^2 x=\sec ^2 x$$
Also, remember the secant is positive in the first quadrant, so
$$\mathrm S =\int_0^{\pi/4} \sqrt{\sec^2 x}dx$$ 
$$\mathrm S =\int_0^{\pi/4} \sec xdx$$ 
A: 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. 
A: The arclength formula should be $\int \sqrt{1+\tan^2 x}\ dx$.
