Reduction formula for $$I_n = \int \tan^n x dx$$

Let $u(x) = \tan^{n-2} x \qquad \qquad v^{'}(x) = \tan^2 x$

Integrating by parts

$$I_n = (\tan x - x)(\tan^{n-2} x) - \int (\tan(x) - x)(n-2)(\tan^{n - 3} x + \tan^{n- 1} x) dx \tag{0}$$

Let the second integral be $J$,

After expanding and simplifying I get,

$$J = (n-2)I_{n-2} + (n-2)I_n - (n-2)\int x\tan^{n-3} x + x\tan^{n-1} dx \tag{1}$$

Let $w(x) = \tan^n x \qquad \qquad z^{'}(x) = 1$

Integrating $I_n$ again by parts,

$$I_n = x\tan^n x - n\int x\tan^{n+1} x + x\tan^{n-1} x dx$$

For $I_{n-2}$,

$$(n-2)\int x\tan^{n-3} x + x\tan^{n-1} dx = x\tan^{n-2} x - I_{n-2}$$

Substituting this in $(1)$

$$J = I_{n-2}(n-1) + I_n(n-2) - x\tan^{n-2} x $$

Substituting this in $(0)$,

we get,

$$I_{n} = {\tan^{n-1} x \over n-1} - I_{n-2}$$.

Is this correct ? I know the final answer is correct but is the method correct ? looks a bit circular to me.


You don't need integration by parts at all. Just write $$\tan^nx=\tan^{n-2}x(\sec^2x-1)$$ Multiply out and integrate and get the answer immediately

  • $\begingroup$ Yes this was described in the book, but is my method correct ? $\endgroup$ – A---B Apr 9 '17 at 4:34
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    $\begingroup$ Yes, your solution is correct, good work! :) $\endgroup$ – Mark Pineau Apr 9 '17 at 4:39
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    $\begingroup$ This doesn't really answer the question, though $\endgroup$ – Robin Aldabanx Apr 9 '17 at 6:07
  • $\begingroup$ @A---B yes your method is essentially correct apart from some typos in line (0) which are subsequently rectified in line (1), but it does seem very laborious. $\endgroup$ – David Quinn Apr 9 '17 at 12:10
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    $\begingroup$ @A---B you are very welcome. Glad to be of help. $\endgroup$ – David Quinn Apr 9 '17 at 16:53

consider $I_n=\int(tanx)^ndx$ and then $I_{n-2}=\int(tanx)^{n-2}dx$ $I_n+I_{n-2}=\int(tanx)^{n-2}(secx)^2dx=\int(u^{n-2})du=\frac{(tanx)^{n-1}}{n-1}$ where $tanx=u$

so the recursion is $I_n+I_{n-2}=\frac{(tanx)^{n-1}}{n-1}$


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