# Improper integrals in curve length

I am supposed to find the length of curve of the following:

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

2. $y =\ln(\cos x)$; $0\le x\le \frac{\pi}{3}$

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:

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

2. $\ell = \int\limits_{0}^{\pi/3} \sqrt{1+(\frac{-\sin x}{\cos x})^2} \ dx$

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!

• For the first, simplify the integrand to $\frac{\sqrt{2}}{\sqrt{2-x^2}}$. Then make substitution $x=t\sqrt{2}$ and recognize the answer, or let $x=\sqrt{2}\sin\theta$. – André Nicolas Apr 8 '14 at 20:00
• @AndréNicolas Sorry, i don't understand on how to simplify to achieve $\frac{\sqrt{2}}{\sqrt{2-x^2}}$. Could you explain? :) – Phantom Apr 8 '14 at 20:36
• @Phantom Use Algebra: $$\sqrt{1+\frac{x^2}{2-x^2}}=\sqrt{\frac{2-x^2}{2-x^2}+\frac{x^2}{2-x^2}}=\sqrt{\frac{2-x^2+x^2}{2-x^2}}=\sqrt{\frac{2}{2-x^2}}=\frac{\sqrt{2}}{\sqrt{2-x^2}}.$$ – Américo Tavares Apr 8 '14 at 20:50
• @AméricoTavares Thank you! I have managed to solve and got this answer: $\sqrt{2}arcsin(\frac{x}{\sqrt2}) + c$ – Phantom Apr 8 '14 at 21:11
• @Phantom This is the right answer, but you need to calculate definite integral, that is take it in the limits from $0$ to $1$. The final answer should be $$\sqrt{2}\arcsin\frac{1}{\sqrt{2}} = \frac{\pi \sqrt{2}}{4}$$ – Adam Latosiński Jun 22 '19 at 23:17

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?
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 $$\sec x$$.