`int_(0)^(pi//2) ""(sin x - cos x)/( 1-sin x * cos x) dx ` is equal to

Options

A. 0
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `pi`

Explanation

To solve the integral

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

we can use a symmetry property of definite integrals.

### Step 1: Use the symmetry property
We know that

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

For our case, \( a = 0 \) and \( b = \frac{\pi}{2} \). Thus, we can rewrite the integral as:

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

### Step 2: Simplify the expression
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 two 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. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x - \sin x}{1 - \cos x \sin x} \, dx \)

Adding these two equations:

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

### Step 4: Simplify the sum
The two fractions have the same denominator, so we can combine them:

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

The numerator simplifies to zero:

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

Thus, we have:

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

### Step 5: Solve for \( I \)
From \( 2I = 0 \), we conclude that:

\[ I = 0. \]

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

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