Question

Solve the integral: (1-x^2)^(3/2)/x^2

This is what I have so far. I used trig substitution.
x=sin theta
dx=cos theta dtheta

[1-(sin^2 theta)^(3/2)]*[cos theta dtheta]/(sin^2 theta)
[(cos^3 theta)]*[cos theta dtheta]/(sin^2 theta)
(cos^4 theta)/(sin^2 theta)

Answer 1

so far so good
cos^4θ/sin^2θ = (1-2sin^2θ+sin^4θ)/sin^2θ
= csc^2θ - 2 + sin^2θ
= csc^2θ - 2 + (1-cos2θ)/2

all of those are simple to integrate.

Answer 2

Steve thx for answering at that time