Use trigonometric substitution. We have an integral with a square root of the form #sqrt(a^2 - x^2)#. Therefore, use the substitution #x = asintheta#.
In our case, #a = 2#. Our substitution will therefore be #x = 2sintheta#. Then #dx- 2costhetad theta#.
#=>intsqrt(4 - (2sintheta)^2)/(2sintheta)^2 * 2costheta d theta#
#=>intsqrt(4 - 4sin^2theta)/(4sin^2theta) * 2costheta d theta#
#=>intsqrt(4(1 - sin^2theta))/(4sin^2theta) * 2costheta d theta#
Use #sin^2x + cos^2x = 1#:
#=>intsqrt(4cos^2theta)/(4sin^2theta) * 2costheta d theta#
#=>int(2costheta)/(4sin^2theta) * 2costheta d theta#
#=>int (4cos^2theta)/(4sin^2theta) d theta#
#=>int cos^2theta/sin^2theta d theta#
Use #cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx#:
#=>int cot^2theta d theta#
Use #cot^2x + 1 = csc^2x#:
#=>intcsc^2theta - 1 d theta#
#=>intcsc^2theta d theta- int 1 d theta#
These are both trivial integrals.
#=>-cottheta - theta + C#
We know from our original substitution that #x/2 = sintheta#. We can deduce that #theta = arcsin(x/2)#. Therefore, the side adjacent #theta# is #sqrt(4 - x^2)# in measure. Thus, #cottheta = sqrt(4 - x^2)/x#.
#=>-sqrt(4 -x^2)/x - arcsin(x/2) + C#
Hopefully this helps!
