determine the following indefinite integral.\n\\(\\int(-7\\sec x\\tan x - 9\\sec^{2}x)dx\\)\n\\(\\int(-7\\sec…

determine the following indefinite integral.\n\\(\\int(-7\\sec x\\tan x - 9\\sec^{2}x)dx\\)\n\\(\\int(-7\\sec x\\tan x - 9\\sec^{2}x)dx = \\square\\)
Answer
Explanation:
Step1: Use integral sum - difference rule
$\int(- 7\sec x\tan x - 9\sec^{2}x)dx=\int(-7\sec x\tan x)dx+\int(-9\sec^{2}x)dx$
Step2: Factor out constants
$=-7\int\sec x\tan xdx-9\int\sec^{2}x dx$
Step3: Recall integral formulas
We know that $\int\sec x\tan xdx=\sec x + C_1$ and $\int\sec^{2}x dx=\tan x + C_2$. So, $-7\int\sec x\tan xdx-9\int\sec^{2}x dx=-7\sec x-9\tan x + C$ (where $C = C_1 + C_2$)
Answer:
$-7\sec x-9\tan x + C$