The problem was to find $\int \tan^5(x)\, \sec^7(x)\, dx$
The solution the book got was different from mine, but I'm sure I did the right steps. Since the solutions are completely different, I have a feeling they might be the same thing, but I'm not sure how I would know this. Here's what the book got:
$\sec^{11}(x)/11-2\sec^9(x)/9+\sec^7(x)/7+C$
Here's what I got: $\tan^6(x)/6 + \tan^{11}(x)/11 + C$
EDIT: So my answer is wrong then, so let me list my steps see if you can catch my mistake
$\int \tan^5(x)\sec^7(x)\, dx$
$\int \tan^4x \tan x \sec^5x \sec^2x\, dx$
$\int \tan^4 x\tan x \ (1+\tan^5x)\,\sec^2x\,dx$
$u=\tan x \ du=\sec^2x$
$\int u^4(u)(1+u^5)\,du$
$\int u^5(1+u^5)\,du$
$\int u^5+u^{10}\,du$
$=\tan^6(x)/6 + \tan^{11}(x)/11 + C$