Using properties of definite integrals, evaluate:
`int_(0)^(pi//2) (sin x - cos x)/(1+sin x cos x) dx`

Answer

To evaluate the integral

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

we can use the properties of definite integrals.

### Step 1: Apply the property of definite integrals
We know that

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). So we can write:

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

### Step 2: Simplify the integrand
Using the identities \( \sin\left(\frac{\pi}{2} - x\right) = \cos x \) and \( \cos\left(\frac{\pi}{2} - x\right) = \sin x \), we can rewrite the integral:

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

### Step 3: Combine the integrals
Now we have two expressions for \( I \):

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

Adding these two equations gives:

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

### Step 4: Simplify the sum
The terms in the integrand cancel out:

\[ \sin x - \cos x + \cos x - \sin x = 0. \]

Thus, we have:

\[ 2I = \int_{0}^{\frac{\pi}{2}} 0 \, dx = 0. \]

### Step 5: Solve for \( I \)
This implies:

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

### Final Result
The value of the integral is:

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