Integrate the following :
`int(1)/(1-sin x)dx`

Explanation

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

### Step 1: Multiply the numerator and denominator by \( 1 + \sin x \)

We start by rewriting the integral:

\[ \int \frac{1}{1 - \sin x} \, dx = \int \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} \, dx \]

### Step 2: Simplify the denominator

Using the identity \( (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x \), we can rewrite the integral:

\[ \int \frac{1 + \sin x}{\cos^2 x} \, dx \]

### Step 3: Split the integral

Now we can split the integral into two parts:

\[ \int \frac{1}{\cos^2 x} \, dx + \int \frac{\sin x}{\cos^2 x} \, dx \]

### Step 4: Integrate each part

The first integral \( \int \frac{1}{\cos^2 x} \, dx \) is known to be:

\[ \int \sec^2 x \, dx = \tan x + C_1 \]

The second integral can be solved using the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \). Thus, we have:

\[ \int \frac{\sin x}{\cos^2 x} \, dx = -\int \frac{1}{u^2} \, du = \frac{1}{u} + C_2 = \frac{1}{\cos x} + C_2 = \sec x + C_2 \]

### Step 5: Combine the results

Combining both parts, we get:

\[ \int \frac{1}{1 - \sin x} \, dx = \tan x + \sec x + C \]

where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer

Thus, the final result is:

\[ \int \frac{1}{1 - \sin x} \, dx = \tan x + \sec x + C \]

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