Evaluate `int(cosx-sinx)/(cosx+sinx)(2+2sin2x)dx`

Explanation

To evaluate the integral

\[ \int \frac{\cos x - \sin x}{(\cos x + \sin x)(2 + 2\sin 2x)} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integral
First, we can factor out the constant from the denominator:

\[ 2 + 2\sin 2x = 2(1 + \sin 2x). \]

Thus, we can rewrite the integral as:

\[ \int \frac{\cos x - \sin x}{(\cos x + \sin x)(2(1 + \sin 2x))} \, dx = \frac{1}{2} \int \frac{\cos x - \sin x}{(\cos x + \sin x)(1 + \sin 2x)} \, dx. \]

### Step 2: Use the identity for \(\sin 2x\)
Recall that \(\sin 2x = 2\sin x \cos x\). Therefore, we can rewrite \(1 + \sin 2x\) as:

\[ 1 + \sin 2x = 1 + 2\sin x \cos x. \]

### Step 3: Rewrite the integral
Now we can express the integral as:

\[ \frac{1}{2} \int \frac{\cos x - \sin x}{(\cos x + \sin x)(1 + 2\sin x \cos x)} \, dx. \]

### Step 4: Substitute \(u = \cos x + \sin x\)
Let \(u = \cos x + \sin x\). Then, the derivative \(du\) is:

\[ du = (-\sin x + \cos x) \, dx. \]

This means that \(\cos x - \sin x = du\).

### Step 5: Change the variable in the integral
Now, we can change the variable in the integral:

\[ \frac{1}{2} \int \frac{du}{u(1 + 2\sin x \cos x)}. \]

### Step 6: Express \(\sin x \cos x\) in terms of \(u\)
Using the identity \(\sin^2 x + \cos^2 x = 1\), we can express \(\sin x \cos x\) as:

\[ \sin x \cos x = \frac{1}{2} \sin 2x. \]

### Step 7: Solve the integral
Now we can integrate:

\[ \frac{1}{2} \int \frac{du}{u(1 + \sin 2x)}. \]

This integral can be solved using partial fractions or logarithmic integration techniques.

### Step 8: Final result
After integrating and substituting back, we find:

\[ \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |\cos x + \sin x| + C. \]

### Conclusion
Thus, the final answer is:

\[ \frac{1}{2} \ln |\cos x + \sin x| + C. \]