To solve the integral
\[
I = \int_0^{\frac{\pi}{4}} \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx,
\]
we can start by rewriting the integrand.
### Step 1: Rewrite the integrand
We know that \(\tan x = \frac{\sin x}{\cos x}\), so we can express \(\sqrt{\tan x}\) as \(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}\). Thus, we can rewrite the integral as:
\[
I = \int_0^{\frac{\pi}{4}} \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int_0^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sqrt{\cos x} \sin x \cos x} \, dx = \int_0^{\frac{\pi}{4}} \frac{\sqrt{\sin x}}{\sin x \cos^2 x} \, dx.
\]
### Step 2: Simplify the integrand
Now, we can simplify the integrand further:
\[
I = \int_0^{\frac{\pi}{4}} \frac{1}{\cos^2 x} \cdot \frac{1}{\sqrt{\cos x}} \, dx = \int_0^{\frac{\pi}{4}} \sec^2 x \cdot \frac{1}{\sqrt{\cos x}} \, dx.
\]
### Step 3: Use substitution
Next, we can use the substitution \(t = \tan x\). Then, \(dt = \sec^2 x \, dx\), and we need to 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 integral becomes:
\[
I = \int_0^1 \frac{1}{\sqrt{1 + t^2}} \, dt.
\]
### Step 4: Evaluate the integral
The integral \(\int \frac{1}{\sqrt{1 + t^2}} \, dt\) is known to be \(\sinh^{-1}(t)\) or \(\ln(t + \sqrt{1 + t^2})\). Therefore, we can evaluate:
\[
I = \left[ \ln(t + \sqrt{1 + t^2}) \right]_0^1.
\]
Calculating the limits:
- At \(t = 1\):
\[
\ln(1 + \sqrt{1 + 1^2}) = \ln(1 + \sqrt{2}).
\]
- At \(t = 0\):
\[
\ln(0 + \sqrt{1 + 0^2}) = \ln(1) = 0.
\]
Thus, we have:
\[
I = \ln(1 + \sqrt{2}) - 0 = \ln(1 + \sqrt{2}).
\]
### Final Result
Therefore, the value of the integral is:
\[
I = \ln(1 + \sqrt{2}).
\]
