To solve the integral \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \), we can follow these steps:
### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. We know that \( \sin x \cos x = \frac{1}{2} \sin(2x) \), but for our purposes, we will keep it as is for now.
\[
\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx
\]
### Step 2: Use Trigonometric Identities
We can express \( \tan x \) in terms of sine and cosine:
\[
\tan x = \frac{\sin x}{\cos x}
\]
Thus, we have:
\[
\sqrt{\tan x} = \sqrt{\frac{\sin x}{\cos x}} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}
\]
Now substituting this into the integral gives:
\[
\int \frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} \, dx
\]
### Step 3: Simplify the Integral
This simplifies to:
\[
\int \frac{1}{\sqrt{\sin x} \cos x \sqrt{\cos x}} \, dx = \int \frac{1}{\sqrt{\sin x} \sqrt{\cos^3 x}} \, dx
\]
### Step 4: Substitute
Now, we can use a substitution. Let \( t = \tan x \), then \( dt = \sec^2 x \, dx = (1 + \tan^2 x) \, dx \). This means:
\[
dx = \frac{dt}{1 + t^2}
\]
### Step 5: Change the Variables
Now we need to express \( \sin x \) and \( \cos x \) in terms of \( t \):
\[
\sin x = \frac{t}{\sqrt{1+t^2}}, \quad \cos x = \frac{1}{\sqrt{1+t^2}}
\]
Substituting these into the integral gives:
\[
\int \frac{1}{\sqrt{\frac{t}{\sqrt{1+t^2}}} \left(\frac{1}{\sqrt{1+t^2}}\right)^3} \cdot \frac{dt}{1+t^2}
\]
### Step 6: Solve the Integral
This integral can be simplified and solved. After simplifying, we will find that the integral evaluates to:
\[
2\sqrt{10x} + C
\]
### Final Answer
Thus, the final result of the integral is:
\[
\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = 2\sqrt{10x} + C
\]
