As with most problems of this type, we'll solve it using a conjugate multiplication trick. Whenever you have something divided by something plus/minus something (as in 1/(cosx-1)), it's always helpful to try conjugate multiplication, especially with trig functions.
We will begin by multiplying 1/(cosx-1) by the conjugate of cosx-1, which is cosx+1:
1/(cosx-1)*(cosx+1)/(cosx+1)
You may wonder why we do this. It's so we can apply the difference of squares property, (a-b)(a+b)=a^2-b^2, in the denominator, to simplify it a little. Back to the problem:
1/(cosx-1)*(cosx+1)/(cosx+1)=(cosx+1)/((cosx-1)(cosx+1))
(underbrace(cosx)-underbrace(1))(underbrace(cosx)+underbrace1))
color(white)(III)acolor(white)(XXX)bcolor(white)(XXX)acolor(white)(XXX)b
Notice how this is essentially (a-b)(a+b).
=(cosx+1)/(cos^2x-1)
Now, what about cos^2x-1? Well, we know sin^2x=1-cos^2x. Let's multiply that by -1 and see what we get:
-1(sin^2x=1-cos^2x)->-sin^2x=-1+cos^2x
=cos^2-1
It turns out that -sin^2x=cos^2x-1, so let's replace cos^2x-1:
(cosx+1)/(-sin^2x
This is equivalent to cosx/-sin^2x+1/-sin^2x, which, using some trig, boils down to -cotxcscx-csc^2x.
At this point, we've simplified to integral int1/(cosx-1)dx to int-cotxcscx-csc^2xdx. Using the sum rule, this becomes:
int-cotxcscxdx+int-csc^2xdx
The first of these is cscx (because the derivative of cscx is -cotxcscx) and the second is cotx (because the derivative of cotx is -csc^2x). Add on the constant of integration C and you have your solution:
int1/(cosx-1)dx=cscx+cotx+C
