I Understanding Reduction Formula

[chwala]

TL;DR

To check on how to arrive at $I_n = \dfrac{1}{n-1}⋅ \tan^{n-1} x - I_{n-2}$

I am looking at $\int \tan^n dx$ where $n$ is a positive integer. The index $n$ has been reduced by writing $\tan^n x$ as $\tan ^{n-2} \tan^2 x$ which is quite clear with me.

We have,

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

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

now this is the part that i need some insight how they moved to

$I_n = \dfrac{1}{n-1}⋅ \tan^{n-1} x - I_{n-2}$

not to say that i do not get it...i want to check if i am doing it right,

My steps,

given,

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

i let $u = tan x$ therefore $u^{'} = \sec^2 x$ and writing $\int \tan^n x dx$ as $I_n$,

$I_n = \int u^{n-2} du - I_{n-2}$

$I_n = \dfrac{1}{n-1}⋅ u^{n-1} - I_{n-2}$

$I_n = \dfrac{1}{n-1}⋅ \tan^{n-1} x - I_{n-2}$

Thanks and cheers.

 

[fresh_42]

I haven't checked but I think you should use $\dfrac{d}{dx}\tan x = 1+\tan^2 x$ to remain in the tangent regime.

 

[Gavran]

Both ways
$$ \begin{align}
&\frac{d}{dx}\tan x=\sec^2x\\
&\frac{d}{dx}\tan x=(1+\tan^2x)
\end{align} $$
produce the same result and they are almost the same. Which one will be used will probably depend on the habit.

 

[chwala]

Both ways
$$ \begin{align}
&\frac{d}{dx}\tan x=\sec^2x\\
&\frac{d}{dx}\tan x=(1+\tan^2x)
\end{align} $$
produce the same result and they are almost the same. Which one will be used will probably depend on the habit.

I was guessing so...