Improper integrals in curve length I am supposed to find the length of curve of the following:


*

*$ y = \sqrt{2-x^2}$ ;  $0\le  x\le 1$

*$y =\ln(\cos x) $; $0\le x\le \frac{\pi}{3}$
I followed the directions found from this question : Length of a curve y = 1 - √x to solve till the integral. So currently i have this for the 2 questions:


*

*$\ell = \int\limits_{0}^{1} \sqrt{1+\frac{x^2}{2-x^2}} \ dx $

*$\ell = \int\limits_{0}^{\pi/3} \sqrt{1+(\frac{-\sin x}{\cos x})^2} \ dx $
However, i am having difficulties integrating the integrals and evaluating due to the square root. Can someone guide me in integrating the integrals above?
All help and suggestions are appreciated. Thank you!

In comments you've said that you've already managed to calculate the first integral, so let's talk about the second.
We have $$ \sqrt{1+\left(\frac{-\sin x}{\cos x}\right)^2} = \sqrt{1+\frac{\sin^2 x}{\cos^2 x}} =\sqrt{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}} = \frac{1}{|\cos x|}$$
For $x\in(0,\frac\pi 3)$ we have $\cos x > 0$, so the integral to be calculated is $$ \ell = \int_0^{\frac\pi 3}\frac{dx}{\cos x}$$
The standard substitution for similar integrals is $t=\tan\frac{x}{2}$, $x = 2\arctan t$, which gives $$ dx = \frac{2\,dt}{1+ t^2}$$ $$\cos x = \cos^2\frac x 2 - \sin^2\frac x 2 = \frac{\cos^2\frac x 2 - \sin^2\frac x 2 }{\cos^2\frac x 2 + \sin^2\frac x 2 } = \frac{1 - \tan^2\frac x 2 }{1 + \tan^2\frac x 2 } = \frac{1-t^2}{1+t^2}$$
$$t_1 = \tan 0 = 0, \qquad t_2 = \tan\frac\pi 6 = \frac{1}{\sqrt{3}}$$
so the integral to calculate is 
$$ \ell = \int_0^{1/\sqrt{3}} \frac{2 dt}{1-t^2}$$
Can you continue?
