How do you find the integral of ## 1/sin^2(x)##?

Maybe more "intuitive" instead of remembering :

##int1/sin^2(x) dx=int (1/cos^2(x))/(sin^2(x)/cos^2(x))dx = int (1/cos^2(x))/tan^2(x)dx##

let's ##u = tan(x)##

##du = 1/cos^2(x) dx##

##int 1/u^2du ##

which is

##[-1/u]##

and remember that u = tan(x) ::

##[-1/tan(x)]##