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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.

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2 Answers 2

up vote 2 down vote accepted

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

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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.

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