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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 2

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

  • 1
    Yes @H.R. that's it! More exactly it is $(-1)^n I_0$ and $(-1)^n I_1$ (because of the change of sign at every iteration). Once you got $I_{n} + I_{n-2}=\frac{1}{n-1}\tan(x)^{n-1}+C_n$ simply rewrite this for $n\to 2n$ as $I_{2n}= \frac{1}{2n-1}\tan(x)^{2n-1}-I_{2n-2}+C_{2n}$ getting thus(applying the recursion to $I_{2n-2}$) $I_{2n}= \frac{1}{2n-1}\tan(x)^{2n-1}-(\frac{1}{2n-2-1}\tan(x)^{2n-3}-I_{2n-4}+C_{2n-2})+C_{2n}=\cdots$ (stopping at $(-1)^n I_{2n-2n}=(-1)^n I_0$).2016-03-18
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I think that you can omit the constant C for the last two equations because the left part can take in the constant.