#intsqrt(4-x^2)/xdx#
Let's use the substitution #x=2sintheta#. This substitution is motivated by the fact that #sqrt(4-x^2)=sqrt(4-4sin^2theta)=2sqrt(1-sin^2theta)=2costheta#. This also implies that #dx=2costhetad theta#.
Using these, the integral becomes:
#=int(2costheta)/(2sintheta)(2costhetad theta)=2intcos^2theta/sinthetad theta=2int(1-sin^2theta)/sinthetad theta#
#=2int(csctheta-sintheta)d theta#
These are both well-known integrals:
#=2(-lnabs(csctheta+cottheta)-costheta)+C#
We have to return to the variable #x#. Recall that #sintheta=x/2# , so we have a right triangle where the side opposite #theta# is #x#, the hypotenuse is #2#, and then by the Pythagorean theorem the side adjacent #theta# is #sqrt(4-x^2)#.
Thus #csctheta=2/x#, #cottheta=sqrt(4-x^2)/x#, and #costheta=sqrt(4-x^2)/2#.
#=-2lnabs((2+sqrt(4-x^2))/x)-sqrt(4-x^2)+C#
There are a lot of different ways to rewrite the natural logarithm here, but this is fine.
