1. Show that

   $$\frac{\cos x}{(1-\sin x)(1+\sin x)}=\sec x$$

2. Using the substitution $u=1-\sin x,$ show that

   $$\int \sec xdx=\int \frac{1}{u(u-2)}du$$

3. Use the method of partial fractions to show that

   $$\int \sec xdx=\ln |\sec x+\tan x|+c$$

If $u=1-\sin (x),$ what is $2-u$?

You should get to the point where you have

Try and multiply the fraction on top and bottom by $1+\sin (x),$ and remember that