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Calculate $\int \tan(x)dx$, $\int \tan^2(x)dx$. Give a formula to $\int \tan^n(x)dx$ in terms of $\int \tan^{n-2}(x)dx$. Use this to calculate $\int \tan^4(x)dx$, $\int \tan^5(x)dx$

I have calculated the first two integrals:

$$\int \tan(x)dx = \log(|\cos(x)|) + C$$ $$\int \tan^2(x)dx = \tan(x) -x + C$$ where $C\in \Bbb R$. But I can't see the relationship between them to obtain the formula. Could you give any hint?

Thanks in advance.

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    $\begingroup$ There doesn't need to be a relationship between the first two integrals to find a recurrence relation between the $n$th and $(n-2)$nd integrals. $\endgroup$ – Peter Foreman Jan 18 at 20:37
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Use

$$\tan^n x = \tan^{n-2} x(\sec^2x-1) = \tan^{n-2} x(\tan x)’-\tan^{n-2}x$$

to express

$$\int \tan^{n}x dx = \int \tan^{n-2}x \>d(\tan x) -\int \tan^{n-2}xdx = \frac {\tan^{n-1} x}{n-1} - \int \tan^{n-2}xdx$$

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