First of all, let's rewrite the function to integrate in a more simple way: remembering that \sec(x)=1/\cos(x), and of course that \tan(x)=\sin(x)/{\cos(x)} we write the integrand as
\frac{\frac{1}{\cos(x)}\frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos^2(x)}-\frac{1}{\cos(x)}}, which we can simplify into {{\sin(x)}/{\cos^2(x)}}/{{1-\cos(x)}/{\cos^2(x)}}, and finally obtain \frac{-\sin(x)}{\cos(x)-1}.
To integrate this function, we'll use a couple of substitutions: first of all, by choosing t=\cos(x), one has dt=-\sin(x)\ dx, and so
\int \frac{-\sin(x)}{\cos(x)-1} dx becomes
\int \frac{dt}{t-1}.
By choosing y=t-1, one obtains dy=dt, and the integral becomes simply
\int \frac{dy}{y}=\log(y)+c. Substituting back y=t-1, one has \log(t-1)+c, and again, plugging t=\cos(x) into the equation, one has \log(\cos(x)-1)+c.
By deriving, you can check that, infact, the following equation holds:
d/dx \log(\cos(x)-1)+c = \frac{-\sin(x)}{\cos(x)-1}
WolframAlpha for checking the derivative
