Integrating an equation with both cos and tan $$\int2\cos^5x\cdot\tan^6x\cdot dx$$
$$2\int\cos^5x\cdot\frac{\sin^6x}{\cos^6x}\cdot dx$$
$$2\int \frac{\sin^6x}{\cos{x}} dx$$
$$2\int\cos^{-2}x\cdot \sin^6x\cdot \cos{x}\cdot dx$$
$$2\int(1-\sin^2x)^{-1}\cdot \sin^6x\cdot \cos{x}\cdot dx$$
$$u = \sin{x}, du = \cos{x}\cdot dx$$
$$2\int(1-u^2)^{-1}\cdot u^6\cdot du$$
$$2\int\frac{u^6}{1-u^2}\cdot du$$
..and now I'm stuck. Maybe I could do something with long division and partial fractions, but this section (I'm doing review) comes before that one. I feel like I'm missing something horribly obvious!

What you have done seems fine... $$\int \dfrac{\sin^6(x)}{\cos(x)}dx = \int \dfrac{\sin^6(x)}{\cos^2(x)}\cos(x)dx = \int \dfrac{\sin^6(x)}{1-\sin^2(x)}\cos(x)dx = \int \dfrac{u^6 du}{1-u^2}$$
We now have
$$\int \dfrac{u^6 du}{1-u^2} = \int \dfrac{du}{1-u^2} - \int \dfrac{1 - u^6}{1-u^2}du = \dfrac12\int \dfrac{du}{1+u} + \dfrac12\int \dfrac{du}{1-u} - \int(1+u^2+u^4)du$$
I trust you can now integrate this.
