Evaluate the following integrals:
`int (1)/(x cos alpha + 1)dx`

Answer

To evaluate the integral \( I = \int \frac{1}{x \cos \alpha + 1} \, dx \), we will follow these steps:

### Step 1: Substitution
Let us substitute \( t = x \cos \alpha + 1 \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = \cos \alpha \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos \alpha} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ I = \int \frac{1}{t} \cdot \frac{dt}{\cos \alpha} \]

### Step 3: Factor Out the Constant
Since \( \frac{1}{\cos \alpha} \) is a constant, we can factor it out of the integral:
\[ I = \frac{1}{\cos \alpha} \int \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \( \frac{1}{t} \) is \( \log |t| \):
\[ I = \frac{1}{\cos \alpha} \log |t| + C \]

### Step 5: Substitute Back
Now, we substitute back \( t = x \cos \alpha + 1 \):
\[ I = \frac{1}{\cos \alpha} \log |x \cos \alpha + 1| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{1}{\cos \alpha} \log |x \cos \alpha + 1| + C \]

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