# Thread: evaluate the integral with inverse trig func.

1. ## evaluate the integral with inverse trig func.

4ds/ (√4-s 2 )

(integrate from 0 to 1

i tried moving the top 4 outside (as 1/4):
[1/4 sin^-1 (s/2) ] from 0 to 1
1/4 sin^/1 (1/2) = 1/4 (pi/6)= pi/24

where did i go wrong?

2. Originally Posted by genlovesmusic09
4ds/ (√4-s 2 )

(integrate from 0 to 1

i tried moving the top 4 outside (as 1/4):
[1/4 sin^-1 (s/2) ] from 0 to 1
1/4 sin^/1 (1/2) = 1/4 (pi/6)= pi/24

$\int_0^1\frac{4}{\sqrt{4-s^2}}ds=4\int_0^1\frac{1}{\sqrt{4-s^2}}ds=4\int_0^1\frac{1}{\sqrt{(2)^2-s^2}}ds$.
$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin{\frac{x}{a}}+C$