I am trying to find the integral of $\int \tan^4 x \, \sec^6 x dx$
I tried to rewrite as trig identities using $\sec^2 - \tan^2 = 1$ but that got me nowhere so I wrote it like this.
$\int \frac{\sin^4x}{\cos^4 x} \frac{1}{\cos^6 x} dx$
$\int \frac{\sin^4x}{\cos^{10} x} dx$
Then I use the idea that making u substitutions for cos will get rid of a power of sin so I just say that the power will go $4 3 2 1 0$ and I will get a $- + - + -$ sign change. I am not sure if this is correct or really how this works exactly but I did a few steps of it and it seemed to work correctly.
$-1\int \frac{1}{u^{10}} dx$
$-1\int u^{-10} dx$
$-1 \times \frac{u^{-9}}{-9}$
$ \frac{u^{-9}}{9}$
$ \frac{\cos^{-9}}{9}$
This is wrong and I am not sure why.