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.
