To evaluate the integral \(\int \frac{\sec x}{1 + \csc x} \, dx\), we will follow a series of steps to simplify and solve the integral.
### Step 1: Rewrite the integral
We start by rewriting \(\sec x\) and \(\csc x\) in terms of sine and cosine:
\[
\sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}
\]
Thus, the integral becomes:
\[
\int \frac{\frac{1}{\cos x}}{1 + \frac{1}{\sin x}} \, dx
\]
### Step 2: Simplify the denominator
The denominator can be simplified:
\[
1 + \frac{1}{\sin x} = \frac{\sin x + 1}{\sin x}
\]
So, we can rewrite the integral as:
\[
\int \frac{1}{\cos x} \cdot \frac{\sin x}{\sin x + 1} \, dx = \int \frac{\sin x}{\cos x (\sin x + 1)} \, dx
\]
### Step 3: Multiply numerator and denominator by \(\cos x\)
To facilitate integration, we multiply the numerator and denominator by \(\cos x\):
\[
\int \frac{\sin x \cos x}{\cos^2 x (\sin x + 1)} \, dx
\]
### Step 4: Change variables
Let \(t = \sin x\). Then, \(dt = \cos x \, dx\). The integral becomes:
\[
\int \frac{t}{(1 - t^2)(1 + t)} \, dt
\]
### Step 5: Use partial fractions
We will express \(\frac{t}{(1 - t^2)(1 + t)}\) using partial fractions:
\[
\frac{t}{(1 - t^2)(1 + t)} = \frac{A}{1 + t} + \frac{B}{1 - t} + \frac{C}{(1 - t)^2}
\]
Multiplying through by the denominator \((1 - t^2)(1 + t)\) and equating coefficients will help us find \(A\), \(B\), and \(C\).
### Step 6: Solve for coefficients
By substituting suitable values for \(t\) (like \(t = -1\), \(t = 1\), and \(t = 0\)), we can find:
- \(A = \frac{1}{4}\)
- \(B = -\frac{1}{2}\)
- \(C = \frac{1}{4}\)
### Step 7: Rewrite the integral
Now, we can rewrite the integral as:
\[
\int \left( \frac{1/4}{1 + t} - \frac{1/2}{1 - t} + \frac{1/4}{1 - t^2} \right) dt
\]
### Step 8: Integrate each term
Integrating each term separately:
1. \(\int \frac{1/4}{1 + t} \, dt = \frac{1}{4} \ln |1 + t|\)
2. \(\int -\frac{1/2}{1 - t} \, dt = -\frac{1}{2} \ln |1 - t|\)
3. \(\int \frac{1/4}{1 - t^2} \, dt = \frac{1}{4} \frac{1}{2} \ln |1 - t| - \frac{1}{4} \frac{1}{2} \ln |1 + t|\)
Combining these results gives:
\[
\frac{1}{4} \ln |1 + t| - \frac{1}{2} \ln |1 - t| + \frac{1}{4} \ln |1 - t| + C
\]
### Step 9: Substitute back
Finally, substituting back \(t = \sin x\):
\[
\frac{1}{4} \ln |1 + \sin x| - \frac{1}{4} \ln |1 - \sin x| + C
\]
### Final Answer
Thus, the final answer for the integral is:
\[
\frac{1}{4} \ln \left( \frac{1 + \sin x}{1 - \sin x} \right) + C
\]
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