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#
