Prove that the length l of an arc given by the parametric equations $x = \theta$ and $y = (\sec\theta)^2 $ from $\theta = 0$ to $\theta = \frac{\pi}{4}$ is given by $l = \ln(1 + \sqrt{2})$.
I have used the formula
$l = \int_0^\frac{\pi}{4} \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta$
So I solved the equation above until
$l = \int_0^\frac{\pi}{4} \sqrt{1 + (2\tan\theta\sec^2\theta)^2} d\theta$
Is the integral above possible to solve?