If $\displaystyle I_{m,n}=\int \sec^{2n}(x)\tan^{m}(x)dx, $ where $m\geq2,n\geq 1$ and $m,n\in\mathbb{N}$
Then prove that $\displaystyle I_{m,n}=\frac{\sec^{2n}(x)\tan^{m-1}(x)}{2n}-\frac{m-1}{2n}I_{m-2,n+1}$
From $\displaystyle I_{m,n}=\int \sec^{2n}(x)\tan^{m}(x)dx$
$\displaystyle I_{m,n}=\int \sec^{2n-2}(x)\tan^{m}(x)\sec^2(x)dx$
Using integration by parts
$\displaystyle I_{m,n}=[\sec^{2n-2}(x)\int \tan^{m}(x)\sec^2(x)dx-\int (\sec^{2n-2}(x))'\int\tan^{m}(x)\sec^2(x)dx$
Put $\tan(x)=t$ and $\sec^2(x)dx=dt$
$\displaystyle I_{m,n}=\sec^{2n-2}(x)\frac{\tan^{m+1}(x)}{m+1}-\frac{2n-2}{m+1}\int\sec^{2n-3}(x)\sec(x)\tan(x)\tan^{m-1}(x)dx$
$\displaystyle I_{m,n}=\frac{\sec^{2n-2}(x)\tan^{m+1}(x)}{m+1}-\frac{2n-2}{m+1}I_{m+2,n-1}$
But I am not getting my original proving expression.please have a look.
Note that:
$$\frac{d}{dx}\sec^{2n}x=2n\sec^{2n-1}x\sec x \tan x =2n\sec^{2n}x\tan x$$ Hence, you have: $$ I_{m,n}=\int \sec^{2n}x\tan^{m}xdx=\frac{1}{2n}\int 2n\sec^{2n}x\tan x\tan^{m-1}xdx= $$ $$=\frac{\sec^{2n}x\tan^{m-1}x}{2n}-\frac{m-1}{2n}\int\sec^{2n}x\tan^{m-2}x\sec^2xdx=$$ $$=\frac{\sec^{2n}x\tan^{m-1}x}{2n}-\frac{m-1}{2n} I_{m-2,n+1} $$