Integrate : `int (1-sinx)/(cos^2x)dx.`

Answer

To solve the integral \(\int \frac{1 - \sin x}{\cos^2 x} \, dx\), we can break it down into simpler parts. Here’s a step-by-step solution:

### Step 1: Rewrite the Integral
We can separate 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: Integrate the First Part
The first integral, \(\int \frac{1}{\cos^2 x} \, dx\), is known to be:
\[ \int \sec^2 x \, dx = \tan x + C_1 \]

### Step 3: Integrate the Second Part
For the second integral, \(\int \frac{\sin x}{\cos^2 x} \, dx\), we can use the substitution method. Let:
\[ u = \cos x \implies du = -\sin x \, dx \implies -du = \sin x \, dx \]
Thus, the integral becomes:
\[ \int \frac{\sin x}{\cos^2 x} \, dx = -\int \frac{1}{u^2} \, du = -\left(-\frac{1}{u}\right) + C_2 = \frac{1}{\cos x} + C_2 = \sec x + C_2 \]

### Step 4: Combine the Results
Now, we can combine the results of both integrals:
\[ \int \frac{1 - \sin x}{\cos^2 x} \, dx = \tan x - \sec x + C \]
where \(C = C_1 + C_2\) is the constant of integration.

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

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