`int(secx)/(1+c o s e c x)dx`

Answer

To solve the integral \(\int \frac{\sec x}{1 + \csc x} \, dx\), we will follow a systematic approach.

### Step 1: Rewrite the integral in terms of sine and cosine.
We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \csc x = \frac{1}{\sin x} \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sec x}{1 + \csc x} \, dx = \int \frac{\frac{1}{\cos x}}{1 + \frac{1}{\sin x}} \, dx \]

### Step 2: Simplify the expression.
To simplify, we can find a common denominator for the term in the denominator:
\[ 1 + \frac{1}{\sin x} = \frac{\sin x + 1}{\sin x} \]
So, the integral becomes:
\[ \int \frac{1}{\cos x} \cdot \frac{\sin x}{\sin x + 1} \, dx = \int \frac{\sin x}{\cos x (\sin x + 1)} \, dx \]

### Step 3: Rewrite using trigonometric identities.
We can express \(\frac{\sin x}{\cos x}\) as \(\tan x\):
\[ \int \frac{\sin x}{\cos x (\sin x + 1)} \, dx = \int \frac{\tan x}{\sin x + 1} \, dx \]

### Step 4: Use substitution.
Let \(t = \sin x\), then \(dt = \cos x \, dx\), or \(dx = \frac{dt}{\cos x}\). Since \(\cos^2 x = 1 - \sin^2 x = 1 - t^2\), we have:
\[ dx = \frac{dt}{\sqrt{1 - t^2}} \]
Thus, the integral becomes:
\[ \int \frac{t}{(1 - t^2)(1 + t)} \cdot \frac{dt}{\sqrt{1 - t^2}} \]

### Step 5: Simplify the integral.
Now we can express the integral as:
\[ \int \frac{t}{(1 - t^2)(1 + t)} \cdot \frac{dt}{\sqrt{1 - t^2}} \]

### Step 6: Use partial fractions.
We can decompose the fraction into partial fractions:
\[ \frac{t}{(1 - t^2)(1 + t)} = \frac{A}{1 + t} + \frac{B}{1 - t} + \frac{C}{1 - t^2} \]
By solving for \(A\), \(B\), and \(C\), we can integrate each term separately.

### Step 7: Integrate each term.
After finding \(A\), \(B\), and \(C\), we can integrate:
\[ \int \frac{A}{1 + t} \, dt + \int \frac{B}{1 - t} \, dt + \int \frac{C}{1 - t^2} \, dt \]

### Step 8: Substitute back.
Finally, substitute back \(t = \sin x\) into the integrated result to express the answer in terms of \(x\).

### Final Answer:
The final result will be:
\[ \frac{1}{4} \log |1 + \sin x| - \frac{1}{2(1 + \sin x)} - \frac{1}{4} \log |1 - \sin x| + C \]