I have to integrate $\frac{1}{1-x^2}dx$ using trig substitution. I know there is a way to do it with partial fractions which yields an answer of $\frac{-1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C$, however I have to show that the trig substitution and partial fractions methods both work.
I know for this problem that $x=\sin\theta, dx=\cos\theta\ d\theta, x^2=\sin^2\theta.$ So now I have $$\int \frac{\cos\theta}{1-\sin^2\theta} d\theta = \int \sec\theta \ d\theta = \ln|sec\theta+\tan\theta|.$$ This is where I have my problem because I don't know what to substitute back in for $\sec\theta$ and $\tan\theta.$