integral problem: formula for int [ tan^k(x) ] dx

[jazzman]

Hi all.
I'm trying to reach a formula that connects between $\displaystyle \tan^{k}{x}$ and $\displaystyle \tan^{k-2}{x}$. I have tried to use integration by parts as such:
$\displaystyle \int{\tan^{k}{x}}=\int{\tan^{2}{x}\cdot\tan^{k-2}{x}}=\left(\tan{x}-x\right)\tan^{k-2}{x}-\int{\left(\tan{x}-x\right)\left(k-2\right)\tan^{k-3}{x}\cdot\frac{1}{cos^{2}{x}}}$
But I'm pretty much stuck there.
Can anyone help me??

 

[pka]

Re: integral problem

Use $\displaystyle \tan ^2 (x) = \sec ^2 (x) - 1$.
That reduces the problem.

 

[jazzman]

Re: integral problem

Use $\displaystyle \tan ^2 (x) = \sec ^2 (x) - 1$.

I don't see how this can help me.
Here is what I got to so far:
$\displaystyle I=\int{tan^{k}{x}}\,dx=(\tan{x}-x)\tan^{k-2}{x}-\int{((k-2)(\tan{x}-x)\tan^{k-3}{x}\cdot\sec^2{x})}\,dx=$
$\displaystyle =\tan^{k-1}{x}-x\cdot\tan^{k-2}{x}-(k-2)\int{\tan^{k-2}{x}\cdot\sec^2{x}}\,dx+(k-2)\int{x\cdot\tan^{k-3}{x}\cdot\sec^2{x}}\,dx=$
$\displaystyle =\tan^{k-1}{x}-x\cdot\tan^{k-2}{x}-(k-2)\tan^{k-1}{x}+(k-2)(x\cdot\tan^{k-2}{x}-I)$
It seems I'm getting closer but I still can't seem to find the solution...
Please help!

 

[pka]

\(\displaystyle \[\begin{array}{rcl} \int {\tan ^k (x)dx} & = & \int {\sec ^2 (x)\tan ^{k - 2} (x)dx} - \int {\tan ^{k - 2} (x)dx} \\ & = & \frac{1}{{k - 1}}\tan ^{k - 1}(x) - \int {\tan ^{k - 2} (x)dx} \\ \end{array}\)

 

[jazzman]

\(\displaystyle \[\begin{array}{rcl} \int {\tan ^k (x)dx} & = & \int {\sec ^2 (x)\tan ^{k - 2} (x)dx} - \int {\tan ^{k - 2} (x)dx} \\ & = & \frac{1}{{k - 1}}\tan ^{k - 1}(x) - \int {\tan ^{k - 2} (x)dx} \\ \end{array}\)

Thank you so much
Didn't think of opening the brackets there...
I was definitely starting to lose my mind!
Thanks for saving me from that misery! :wink: