1. Stuck: Arc Length

Hello. I need help with a problem.

The problem is: Show that the circumference of the unit circle is equal to

2 (integral, 1, -1) dx/sqrt(1-x^2) (an improper integral). Evaluate, thus verifying that the circumference is 2pi.

This is how far I've gotten:

x^2 + y^2 = 1 --> y = sqrt(1-x^2)

y = sqrt(1-x^2)
y'= -x/sqrt(1-x^2)

(y')^2 + 1 = (-x/sqrt(1-x^2))^2 + 1

= (x^2)/(1-x^2) + 1

Arclength = s = (integral, 1, -1) sqrt(1 + (x^2)/(1-x^2)) dx

= (integral, 1, -1) sqrt(1/(1-x^2)) dx

This is where I am stuck. Thanks!

2. Originally Posted by wyhwang7
Hello. I need help with a problem.

The problem is: Show that the circumference of the unit circle is equal to

2 (integral, 1, -1) dx/sqrt(1-x^2) (an improper integral). Evaluate, thus verifying that the circumference is 2pi.

This is how far I've gotten:

x^2 + y^2 = 1 --> y = sqrt(1-x^2)

y = sqrt(1-x^2)
y'= -x/sqrt(1-x^2)

(y')^2 + 1 = (-x/sqrt(1-x^2))^2 + 1

= (x^2)/(1-x^2) + 1

Arclength = s = (integral, 1, -1) sqrt(1 + (x^2)/(1-x^2)) dx

= (integral, 1, -1) sqrt(1/(1-x^2)) dx

This is where I am stuck. Thanks!
$\sqrt{\frac{1}{1 - x^2}} = \frac{1}{\sqrt{1 - x^2}}$ and I hope you now recognise the integrand as a standard form.