# Computing $I_{n}=\int \tan(x)^n \mathrm dx$

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.

-

## 2 Answers

Yes everything is right. (sorry for the initial answer)

-
OK thank you very much! –  Chon Nov 1 '12 at 18:51

I think that you can omit the constant C for the last two equations because the left part can take in the constant.

-