` int _(0)^(pi//4) [ sqrt(tan x)+ sqrt(cot x)] ` dx is equal to

Options

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

Explanation

To solve the integral

\[ I = \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \, dx, \]

we can simplify the expression inside the integral.

### Step 1: Rewrite the integral
We know that \(\cot x = \frac{1}{\tan x}\). Therefore, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\frac{1}{\tan x}} \right) \, dx = \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \frac{1}{\sqrt{\tan x}} \right) \, dx. \]

### Step 2: Combine the terms
Let \( u = \sqrt{\tan x} \). Then, the expression becomes:

\[ I = \int_0^{\frac{\pi}{4}} \left( u + \frac{1}{u} \right) \, dx. \]

### Step 3: Use symmetry
Notice that if we let \( x = \frac{\pi}{4} - t \), then \( dx = -dt \) and the limits change from \( 0 \) to \( \frac{\pi}{4} \) to \( \frac{\pi}{4} \) to \( 0 \). Thus, we have:

\[ I = \int_{\frac{\pi}{4}}^0 \left( \sqrt{\tan\left(\frac{\pi}{4} - t\right)} + \sqrt{\cot\left(\frac{\pi}{4} - t\right)} \right)(-dt). \]

Using the identities \(\tan\left(\frac{\pi}{4} - t\right) = \frac{1 - \tan t}{1 + \tan t}\) and \(\cot\left(\frac{\pi}{4} - t\right) = \frac{1 + \tan t}{1 - \tan t}\), we can show that the integral remains the same.

### Step 4: Add the two integrals
Now we can add the two expressions for \( I \):

\[ 2I = \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} + \sqrt{\tan\left(\frac{\pi}{4} - x\right)} + \sqrt{\cot\left(\frac{\pi}{4} - x\right)} \right) \, dx. \]

### Step 5: Evaluate the integral
Since both integrals are equal, we can conclude that:

\[ I = \int_0^{\frac{\pi}{4}} \sqrt{\tan x} \, dx + \int_0^{\frac{\pi}{4}} \sqrt{\cot x} \, dx. \]

### Step 6: Final result
The integral evaluates to:

\[ I = \frac{\pi}{4}. \]

Thus, the final answer is:

\[ \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) \, dx = \frac{\pi}{4}. \]