Question

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

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:

$-7\sec x-9\tan x + C$