How do you evaluate $int dx/sqrt(4-x^2)$ from [0,1]?

Question

6877 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-03T21:57:12

$pi/6$

Use the substitution $x = 2sintheta$ . Then $dx = 2costheta d theta$ .

$=>int_0^1 1/sqrt(4 - (2sintheta)^2) * 2costheta d theta$

$=>int_0^1 1/sqrt(4(1 - sin^2theta)) 2costheta d theta$

Apply the pythagorean identity $cos^2x = 1 - sin^2x$ :

$=>int_0^1 1/sqrt(4cos^2theta) 2costheta d theta$

$=>int_0^1 1/(2costheta) 2costheta d theta$

$=>int_0^1 1 d theta$

$=> int_0^1 arcsin(x/2)$

Evaluate using the second fundamental theorem of calculus.

$=>arcsin(1/2) - arcsin(0/2)$

$=> pi/6$

Hopefully this helps!

How do you evaluate int dx/sqrt(4-x^2)int dx/sqrt(4-x^2) from [0,1]?