7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Yes this was described in the book, but is my method correct ? $\endgroup$ – A---B Apr 9 '17 at 4:34
  • 2
    $\begingroup$ Yes, your solution is correct, good work! :) $\endgroup$ – Mark Pineau Apr 9 '17 at 4:39
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @A---B you are very welcome. Glad to be of help. $\endgroup$ – David Quinn Apr 9 '17 at 16:53
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.