# Thread: Reduction method of integration

1. ## Reduction method of integration

Establish a reduction formula for the following integrals.

$\int\tan^n x dx$ (use $tan^2 x\equiv\sec^2 x -1$)

$\int \sec^n x dx$

I figure that both are similar, and that if I can figure out one i should be able to do the other.
The examples i was given would give the method
$\int\tan^n x dx=\int\tan x\tan^{n-1} x dx$ and then use integration by parts, but I'm not sure about this one.
I tried
$\int\tan^n x dx=\int\tan^2 x\tan^{n-2} x dx$
$=\tan^{n-2}x(\tan x -x)-(n-2)\int(\sec^2 x \tan^{n-3} x)(\tan x -x)$
But i could not work it out to the answer which is
$(n-1)(I_n+I_{n-2})=\tan^{n-1} x$, $I_n=\int\tan^n x dx$ and $I_{n-2}=\int\tan^{n-2} x dx$

Thanks!

2. $\displaystyle{\int{\tan^n{x}\,dx} = \int{\tan^2{x}\tan^{n-2}{x}\,dx}}$

$\displaystyle{= \int{(\sec^2{x} - 1)\tan^{n - 2}{x}\,dx}}$

$\displaystyle{= \int{\sec^2{x}\tan^{n - 2}{x} - \tan^{n - 2}{x}\,dx}}$

$\displaystyle{= \int{\tan^{n - 2}{x}\sec^2{x}\,dx} - \int{\tan^{n - 2}{x}\,dx}}$

$\displaystyle{= \int{u^{n - 2}\,du} - \int{\tan^{n - 2}{x}\,dx}}$ after making the substitution $u = \tan{x}$

$\displaystyle{= \frac{u^{n - 1}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}$

$\displaystyle{ = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}$.

So $\displaystyle{\int{\tan^n{x}\,dx} = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 2}{x}\,dx}}$

3. So $\displaystyle{\int{\tan^n{x}\,dx} = \frac{\tan^{n - 1}{x}}{n - 1} - \int{\tan^{n - 1}{x}\,dx}}$
you mean $\displaystyle \int\tan^nx dx=\frac{\tan^{n-1} x}{n-1}-\int\tan^{n-2}x dx$?

4. Yes I do