`int_0^(pi//2) x(sqrt(tan x)+sqrt(cot x))dx` equals

Options

A. `(pi)/(2sqrt(2))`
B. `(pi^(2))/(2)`
C. `(pi^(2))/(2sqrt(2))`
D. `(pi^(2))/(2sqrt(3))`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{2}} x \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \), we can use the symmetry property of definite integrals. Let's go through the steps:

### Step 1: Define the Integral
Let:
\[ I = \int_0^{\frac{\pi}{2}} x \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]

### Step 2: Use the Symmetry Property
We can use the property of definite integrals:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, \( a = \frac{\pi}{2} \). Therefore, we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\tan\left(\frac{\pi}{2} - x\right)} + \sqrt{\cot\left(\frac{\pi}{2} - x\right)} \right) dx \]

### Step 3: Simplify the Functions
Recall that:
\[ \tan\left(\frac{\pi}{2} - x\right) = \cot x \quad \text{and} \quad \cot\left(\frac{\pi}{2} - x\right) = \tan x \]
Thus, we can rewrite the integral:
\[ I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]

### Step 4: Combine the Two Integrals
Now, we can add the two expressions for \( I \):
\[ 2I = \int_0^{\frac{\pi}{2}} x \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx + \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]
This simplifies to:
\[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \right) dx \]

### Step 5: Evaluate the Integral
Now, we can factor out \( \frac{\pi}{2} \):
\[ 2I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]
Thus:
\[ I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]

### Step 6: Evaluate the Integral of \( \sqrt{\tan x} + \sqrt{\cot x} \)
The integral \( \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \) can be evaluated using known results:
\[ \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \, dx = \int_0^{\frac{\pi}{2}} \sqrt{\cot x} \, dx = \frac{\pi}{2} \]
Thus:
\[ \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Step 7: Final Result
Substituting back into our expression for \( I \):
\[ I = \frac{\pi}{4} \cdot \pi = \frac{\pi^2}{4} \]

### Conclusion
The value of the integral is:
\[ \int_0^{\frac{\pi}{2}} x \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx = \frac{\pi^2}{4} \]