Question

How do you evaluate the integral `int 1/(sqrt(4-x^2)dx` from 0 to 2?

Answer 1

`pi/2`.

Explanation:

Knowing that, `int1/sqrt(a^2-x^2)dx=arc sin(x/a)+C`, we have,

`int_0^2 1/sqrt(4-x^2)dx=[arc sin(x/2)]_0^2=arc sin1-arc sin0=pi/2`.