Consider using trigonometric substitution.
Observe we have 4-x^2 It resembles like 1-sin^2(theta) or 1-cos^2(theta)
Another advantage is if we are able to make the given 4-x^2 into one of the two above trigonometric identity we can easily knock out the square root as 1-sin^2(theta) = cos^2(theta) and 1-cos^2(theta) = sin^2(theta)
Let me choose x=sin(theta)
If I do we would get 4-sin^2(theta) which is not going to help!
So I need to substitute for x such a way that it is going to factor out 4.
So ideal choice would be x=2sin(theta)
4-x^2 would become 4-4sin^2(theta)
4-4sin^2(theta) = 4(1-sin^2(theta)) = 4cos^2(theta)
Let us generalize this.
If we have a^2- x^2, we should be substituting x=asin(theta) or x=acos(theta)
Let us move ahead.
x=2sin(theta)
Differentiating with respect to x
dx=2cos(theta) d theta
Our integral int dx/sqrt(4-x^2) after substitution becomes
int (2cos(theta)d theta)/sqrt(4-4sin^2(theta))
=int (2cos(theta) d theta)/sqrt(4(1-sin^2(theta))
=int(2cos(theta) d theta)/sqrt(4cos^2(theta))
=int(2cos(theta) d theta)/(2cos(theta))
=intcancel(2cos(theta) d theta)/cancel(2cos(theta))
=int d theta
= theta + C
We took x = 2sin(theta)
sin(theta)=x/2
theta = sin^-1(x/2)
Our final answer sin^-1(x/2)+C