`int ((1- sinx))/(cos^(2)x) dx= ?`

Options

A. `tanx +sec x + C`
B. `tanx - sec x + C`
C. `-tanx + sec x + C`
D. `-tanx - sec x + C`

Answer

To solve the integral \( \int \frac{1 - \sin x}{\cos^2 x} \, dx \), we can follow these steps:

### Step 1: Split the integral
We can split the integral into two parts:
\[ \int \frac{1 - \sin x}{\cos^2 x} \, dx = \int \frac{1}{\cos^2 x} \, dx - \int \frac{\sin x}{\cos^2 x} \, dx \]

### Step 2: Rewrite the integrals
We know that \( \frac{1}{\cos^2 x} = \sec^2 x \) and \( \frac{\sin x}{\cos^2 x} = \tan x \sec x \). Thus, we can rewrite the integrals:
\[ \int \sec^2 x \, dx - \int \tan x \sec x \, dx \]

### Step 3: Integrate the first term
The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x + C_1 \]

### Step 4: Integrate the second term
The integral of \( \tan x \sec x \) can be solved using the substitution \( u = \tan x \), which gives \( du = \sec^2 x \, dx \). Thus:
\[ \int \tan x \sec x \, dx = \int u \, du = \frac{u^2}{2} + C_2 = \frac{\tan^2 x}{2} + C_2 \]

### Step 5: Combine the results
Now we can combine the results from the two integrals:
\[ \int \frac{1 - \sin x}{\cos^2 x} \, dx = \tan x - \frac{\tan^2 x}{2} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1 - \sin x}{\cos^2 x} \, dx = \tan x - \frac{\tan^2 x}{2} + C \]

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