$$\int \frac{1}{\cos x}dx$$
By half angle I did:
$t = \tan\frac{x}{2}$
$$\cos x = 1-2\sin^2\frac{x}{2} = 1-2\frac{\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}\cos^2\frac{x}{2} = 1-2t^2\frac{1}{\sec^2\frac{x}{2}} =1 - \frac{1-2t^2}{t^2+1} = \frac{1-t^2}{t^2+1}$$
$$dt = \frac{1}{2}\sec^2\frac{x}{2}\implies dx = \frac{2}{1+t^2}dt$$
then $$\int\frac{1}{\cos x} = \int\frac{t^2+1}{1-t^2}\frac{2}{1+t^2}dt = \int\frac{2t^2+2}{1-(t^2)^2}dt$$
I know how to separate it in two integrals:
$$\int\frac{2t^2}{1-(t^2)^2}dt + \int \frac{2}{1-(t^2)^2}dt$$
The first one I know how to integrate, but not the last one. Could somebody help me? Also, Am I doing everything alrigt?
os: I know this integral is easier by $u$ substitution...
If you cancel correctly, you end up with:
$$\int\frac{2\,dt}{1-t^2}$$
But: $$\frac{2}{1-t^2}=\frac{1}{1-t} + \frac{1}{1+t}$$