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.