Integral of $\int \tan^4 x \, \sec^6 x dx$ 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.

Here is a relatively easier way. Remember that $\sec^2(x) = 1 + \tan^2(x)$. Hence the integral $$I = \int \tan^4(x) \sec^6(x) dx = \int \tan^4(x) \sec^4(x) \sec^2(x) dx\\ = \int \tan^4(x) \left(1 + \tan^2(x) \right)^2 \sec^2(x) dx $$
Let $\tan(x) = t$. Hence, $\sec^2(x) dx = dt$. Hence, we get that
$$I = \int t^4 \left( 1+t^2 \right)^2 dt = \int t^4 \left( 1+2t^2 +t^4 \right) dt = \int \left(t^4 + 2t^6 + t^8 \right) dt\\
=\dfrac{t^5}{5} + \dfrac{2t^7}{7} + \dfrac{t^9}{9} + C$$
Hence, we get that $$I = \dfrac{\tan^5(x)}{5} + \dfrac{2\tan^7(x)}{7} + \dfrac{\tan^9(x)}{9} + C$$
