`int_(0)^(pi//2)(x)/(sinx+cosx)dx`.

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x} \, dx \), we will use the property of definite integrals that states:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]

### Step 1: Set up the integral
We start with the integral:

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

### Step 2: Apply the property of definite integrals
Using the property mentioned, we can rewrite the integral as:

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

### Step 3: Simplify the integrand
We know that:

\[ \sin\left(\frac{\pi}{2} - x\right) = \cos x \quad \text{and} \quad \cos\left(\frac{\pi}{2} - x\right) = \sin x \]

Thus, we can simplify the denominator:

\[ \sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right) = \cos x + \sin x \]

Now, substituting this back into the integral gives us:

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

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

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

Adding these two equations gives:

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

### Step 5: Solve for \( I \)
Thus, we can express \( I \) as:

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

### Step 6: Calculate the integral
Now, we need to evaluate the integral:

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

We can factor out \( \frac{\pi}{2} \):

\[ I = \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx \]

### Step 7: Evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx \)
We know that:

\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

Thus, we can rewrite the integral:

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)} \, dx \]

This integral evaluates to \( \frac{\pi}{2\sqrt{2}} \).

### Step 8: Substitute back to find \( I \)
Now substituting back, we have:

\[ I = \frac{\pi}{4} \cdot \frac{\pi}{2\sqrt{2}} = \frac{\pi^2}{8\sqrt{2}} \]

### Final Answer
Thus, the final result for the integral is:

\[ I = \frac{\pi^2}{8\sqrt{2}} \]