`int_(0)^(pi//4)(sqrt(tanx)+sqrt(cotx))dx` equals

Options

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

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral in terms of sine and cosine:
\[ I = \int_{0}^{\frac{\pi}{4}} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) dx \]

### Step 2: Simplify the expression
This can be simplified as:
\[ I = \int_{0}^{\frac{\pi}{4}} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) dx \]
Let’s denote the two terms separately:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}} dx + \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx \]

### Step 3: Change of variable
Now, we can make a substitution for the second integral. Let \( u = \frac{\pi}{4} - x \). Then, \( du = -dx \) and when \( x = 0 \), \( u = \frac{\pi}{4} \) and when \( x = \frac{\pi}{4} \), \( u = 0 \). Thus, we have:
\[ \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx = \int_{\frac{\pi}{4}}^{0} \frac{\sqrt{\sin u}}{\sqrt{\cos u}} (-du) = \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\sin u}}{\sqrt{\cos u}} du \]

### Step 4: Combine the integrals
Now, we can combine the two integrals:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}} dx + \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}} dx = 2 \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}} dx \]

### Step 5: Evaluate the integral
Now, we can evaluate:
\[ I = 2 \int_{0}^{\frac{\pi}{4}} \sqrt{\tan x} dx \]

### Step 6: Use a known integral
The integral \( \int \sqrt{\tan x} \, dx \) can be evaluated using known results or numerical methods. However, we can also use symmetry properties or known integral tables to find that:
\[ \int_{0}^{\frac{\pi}{4}} \sqrt{\tan x} \, dx = \frac{\pi}{4 \sqrt{2}} \]
Thus,
\[ I = 2 \cdot \frac{\pi}{4 \sqrt{2}} = \frac{\pi}{2 \sqrt{2}} \]

### Final Result
Therefore, the value of the integral is:
\[ \boxed{\frac{\pi}{2 \sqrt{2}}} \]