fy the following \( \frac{7 \cos x \sin x+9}{\cos ^{2} x} d x \)

We wish to evaluate \[ \int \frac{7 \cos x \sin x + 9}{\cos^2 x}\,dx. \] **Step 1.** Write the integral as the sum of two integrals: \[ \int \frac{7 \cos x \sin x+9}{\cos^2 x}\,dx = \int \frac{7 \cos x \sin x}{\cos^2 x}\,dx + \int \frac{9}{\cos^2 x}\,dx. \] **Step 2.** Simplify each term: - For the first term: \[ \frac{7 \cos x \sin x}{\cos^2 x} = 7\frac{\sin x}{\cos x} = 7\tan x. \] - For the second term: \[ \frac{9}{\cos^2 x} = 9\sec^2 x. \] Thus, the integral becomes: \[ \int 7\tan x\,dx + \int 9\sec^2 x\,dx. \] **Step 3.** Evaluate the integrals: - Recall that \[ \int \tan x\,dx = -\ln|\cos x| + C_1, \] so \[ \int 7\tan x\,dx = -7\ln|\cos x| + C_1. \] - Also, recall that \[ \int \sec^2 x\,dx = \tan x + C_2, \] so \[ \int 9\sec^2 x\,dx = 9\tan x + C_2. \] **Step 4.** Combine the results: \[ \int \frac{7 \cos x \sin x + 9}{\cos^2 x}\,dx = -7\ln|\cos x| + 9\tan x + C, \] where \(C = C_1 + C_2\) is the constant of integration. The final answer is: \[ -7\ln|\cos x| + 9\tan x + C. \]