`int_(0)^(pi//2) ((cos x-sin x))/((1+sin x cos x)) dx`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. 0
D. none

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx, \]

we can use a symmetry property of definite integrals.

### Step 1: Define the integral
Let

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx. \]

### Step 2: Change of variable
Now, we will perform a change of variable by letting \( x = \frac{\pi}{2} - t \). Then, \( dx = -dt \), and the limits change as follows:
- When \( x = 0 \), \( t = \frac{\pi}{2} \).
- When \( x = \frac{\pi}{2} \), \( t = 0 \).

Thus, we have:

\[ I = \int_{\frac{\pi}{2}}^{0} \frac{\cos\left(\frac{\pi}{2} - t\right) - \sin\left(\frac{\pi}{2} - t\right)}{1 + \sin\left(\frac{\pi}{2} - t\right) \cos\left(\frac{\pi}{2} - t\right)} (-dt). \]

### Step 3: Simplifying the integral
Using the trigonometric identities \( \cos\left(\frac{\pi}{2} - t\right) = \sin t \) and \( \sin\left(\frac{\pi}{2} - t\right) = \cos t \), we can rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin t - \cos t}{1 + \cos t \sin t} \, dt. \]

### Step 4: Combine the integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx \)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \cos x \sin x} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos x - \sin x + \sin x - \cos x}{1 + \sin x \cos x} \right) \, dx = \int_{0}^{\frac{\pi}{2}} \frac{0}{1 + \sin x \cos x} \, dx = 0. \]

### Step 5: Solve for \( I \)
Thus, we have:

\[ 2I = 0 \implies I = 0. \]

### Conclusion
The value of the integral is

\[ \boxed{0}. \]