I'm trying to compute:
$ I_{n}=\int \tan(x)^n \mathrm dx$
We have:
$ I_{n}+I_{n-2}=\int (1+\tan(x)^2)\tan(x)^{n-2} \mathrm dx$
$ I_{n}=\frac{1}{n-1}\tan(x)^{n-1}-I_{n-2}+C$
Which gives the formulas:
$ \int \tan(x)^{2n} \mathrm dx= \sum_{k=0}^{n-1} \frac{(-1)^k}{2n-(2k+1)}\tan(x)^{2n-(2k+1)}+(-1)^nx+C$
$ \int \tan(x)^{2n+1} \mathrm dx=\sum_{k=0}^{n-1} \frac{(-1)^k}{2(n-k)}\tan(x)^{2(n-k)}+(-1)^{n+1}\ln(\cos(x))+C$
I would just like to know if these equalities are correct.