`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 a symmetry property of definite integrals. Here’s the step-by-step solution:

### 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
Using the property of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
we can rewrite \( I \):
\[ I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\tan\left(\frac{\pi}{2} - x\right)} + \sqrt{\cot\left(\frac{\pi}{2} - x\right)} \right) dx \]
Since \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \) and \( \cot\left(\frac{\pi}{2} - x\right) = \tan x \), we have:
\[ I = \int_0^{\frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx \]

### Step 3: Combine the Two Integrals
Now we can combine the two expressions for \( I \):
\[ 2I = \int_0^{\frac{\pi}{2}} \left( x + \left( \frac{\pi}{2} - x \right) \right) \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]
This simplifies to:
\[ 2I = \int_0^{\frac{\pi}{2}} \frac{\pi}{2} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]
Thus:
\[ I = \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]

### Step 4: Evaluate the Integral
Next, we need to evaluate:
\[ \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \]
Using the substitution \( t = \tan x \), we have \( dx = \frac{1}{1+t^2} dt \) and the limits change from \( 0 \) to \( \infty \):
\[ \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \, dx = \int_0^{\infty} \frac{\sqrt{t}}{1+t^2} dt \]
And similarly for \( \sqrt{\cot x} \):
\[ \int_0^{\frac{\pi}{2}} \sqrt{\cot x} \, dx = \int_0^{\infty} \frac{1}{\sqrt{t} (1+t^2)} dt \]

### Step 5: Combine the Results
Both integrals yield the same value, so:
\[ \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx = 2 \int_0^{\infty} \frac{\sqrt{t}}{1+t^2} dt \]
The integral \( \int_0^{\infty} \frac{\sqrt{t}}{1+t^2} dt \) can be evaluated using the beta function or gamma function, yielding \( \frac{\pi}{2\sqrt{2}} \).

### Step 6: Final Calculation
Thus:
\[ \int_0^{\frac{\pi}{2}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx = 2 \cdot \frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}} \]
Finally, substituting back into our expression for \( I \):
\[ I = \frac{\pi}{4} \cdot \frac{\pi}{\sqrt{2}} = \frac{\pi^2}{4\sqrt{2}} \]

### Final Answer
\[ I = \frac{\pi^2}{4\sqrt{2}} \]