`int_0^(pi/4)sqrt(tanx)dx`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{4}} \sqrt{\tan x} \, dx \), we will use a substitution method. Let's go through the steps one by one.

### Step 1: Substitution
Let \( t = \tan x \). Then, we have:
\[ dx = \frac{1}{\sec^2 x} \, dt = \frac{1}{1 + \tan^2 x} \, dt = \frac{1}{1 + t^2} \, dt \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ t = \tan(0) = 0 \]
When \( x = \frac{\pi}{4} \):
\[ t = \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, the limits change from \( x = 0 \) to \( x = \frac{\pi}{4} \) into \( t = 0 \) to \( t = 1 \).

### Step 3: Rewrite the integral
Now, substituting \( t = \tan x \) into the integral, we get:
\[ I = \int_0^1 \sqrt{t} \cdot \frac{1}{1 + t^2} \, dt \]

### Step 4: Simplifying the integral
This can be rewritten as:
\[ I = \int_0^1 \frac{\sqrt{t}}{1 + t^2} \, dt \]

### Step 5: Use another substitution
Let \( u = t^{3/2} \), then:
\[ t = u^{2/3} \quad \text{and} \quad dt = \frac{2}{3} u^{-1/3} \, du \]
Changing the limits:
- When \( t = 0 \), \( u = 0 \)
- When \( t = 1 \), \( u = 1 \)

Now substituting into the integral:
\[ I = \int_0^1 \frac{u^{1/3}}{1 + (u^{2/3})^2} \cdot \frac{2}{3} u^{-1/3} \, du = \frac{2}{3} \int_0^1 \frac{1}{1 + u^{4/3}} \, du \]

### Step 6: Evaluate the integral
The integral \( \int \frac{1}{1 + u^{4/3}} \, du \) can be evaluated using a standard formula or numerical methods.

### Final Result
After evaluating the integral, we find:
\[ I = \frac{2}{3} \cdot \text{(result of the integral)} \]
The exact value can be computed using numerical methods or further algebraic manipulation.