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!
 A: 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?
A: For $1$, you should be able to use $\arcsin x=\int\dfrac1{\sqrt{1-x^2}}\operatorname dx$.
For two, you do a little rearranging and just need to integrate $\sec x$.
