Question 886474
<pre>
{{{int((1-cot(x))^2,dx)}}}

                          Square the binomial:

{{{int((1-2cot(x)+cot^2x),dx)}}}

                          Write as the sum of three integrals:

{{{int((1)*dx)}}}{{{""+""}}}{{{int((-2cot(x))*dx)}}}{{{""+""}}}{{{int((cot^2x)*dx)}}}

                          Take out constant in second integral, and use the
                          Pythagorean identity {{{cot^2x=csc^2x-1}}} in the third integral: 


{{{int(dx)}}}{{{""-""}}}{{{2*int((cot(x))*dx)}}}{{{""+""}}}{{{int((csc^2x-1)*dx)}}}

{{{int(dx)}}}{{{""-""}}}{{{2*int((cot(x))*dx)}}}{{{""+""}}}{{{int((csc^2x)*dx)}}}{{{""+""}}}{{{int((-1)*dx)}}}

{{{int(dx)}}}{{{""-""}}}{{{2*int((cot(x))*dx)}}}{{{""+""}}}{{{int((csc^2x)*dx)}}}{{{""-""}}}{{{int(dx)}}}

                          The first and last integrals cancel

{{{-2*int((cot(x))*dx)}}}{{{""+""}}}{{{int((csc^2x)*dx)}}}

                          Use the integral formulas:

                          {{{int((cot(u)),du)=ln(abs (sin(u)) )+C}}} and {{{int((csc^2u),du)=-cot(u)+C}}}

{{{-2*ln(abs (sin(x)) ) + (-cot(x))+C}}}

{{{-2*ln(abs (sin(x)) ) - cot(x)+C}}}

Edwin</pre>