# Recurrent formula

• June 13th 2012, 06:30 AM
wabaffet
Recurrent formula
I have this kind of problem. I have to find a recurrent formula for
I(n)=integrate (tan(x))^n)dx from n=0 to a, and then I need to fine I(0) and I(1).
I wrote it like this
I(n)=int ((tan(x))^(n-1)*(tan(x)))dx and then tried to integrate it partially.

u=(tan(x))^(n-1)
du=(n-1)*((tan(x))^(n-2))/(cos(x))^2
v=int (tan(x)dx)=-log|cos(x)|
But, when I put it back...
I(n)=-tan(x)*log|cos(x)|-(n-1)*int((tan(x))^(n-2)* (log|cos(x)|/(cos(x))^2)
I have this, and I don't know what to do with this new int.
• June 13th 2012, 06:46 AM
Reckoner
Re: Recurrent formula
Quote:

Originally Posted by wabaffet
I have this kind of problem. I have to find a recurrent formula for
I(n)=integrate (tan(x))^n)dx from n=0 to a, and then I need to fine I(0) and I(1).

$I(n) = \int_0^a\tan^nx\,dx$

$I(0) = \int_0^adx$

$I(1) = \int_0^a\tan x\,dx$

You should be able to integrate those. For $n\geq2,$

$I(n) = \int_0^a\tan^2x\tan^{n-2}x\,dx$

$= \int_0^a\left(\sec^2x-1\right)\tan^{n-2}x\,dx$

$= \int_0^a\sec^2x\tan^{n-2}x\,dx - \int_0^a\tan^{n-2}x\,dx$

$= \int_0^a\sec^2x\tan^{n-2}x\,dx - I(n-2)$

Now you should be able to evaluate the integral on the left...
• June 13th 2012, 02:50 PM
wabaffet
Re: Recurrent formula
I did it, it's all ok at the end. Thank you.