To solve the integral \( \int \frac{1}{\sqrt{\sin^3 x \cos x}} \, dx \), we will follow these steps:
### Step 1: Rewrite the Integral
We start with the integral:
\[
\int \frac{1}{\sqrt{\sin^3 x \cos x}} \, dx
\]
To simplify this, we multiply the numerator and denominator by \( \sec^2 x \):
\[
\int \frac{\sec^2 x}{\sqrt{\sin^3 x \cos x} \cdot \sec^2 x} \, dx
\]
This gives us:
\[
\int \frac{\sec^2 x}{\sqrt{\sin^3 x} \cdot \sqrt{\cos^3 x}} \, dx
\]
### Step 2: Simplify the Denominator
The denominator can be rewritten as:
\[
\sqrt{\sin^3 x \cos x} = \sqrt{\sin^3 x} \cdot \sqrt{\cos x}
\]
Thus, we have:
\[
\int \frac{\sec^2 x}{\sqrt{\sin^3 x} \cdot \sqrt{\cos x}} \, dx
\]
Now, we can express \( \sec^2 x \) as \( \frac{1}{\cos^2 x} \):
\[
\int \frac{1}{\cos^2 x \cdot \sqrt{\sin^3 x \cos x}} \, dx
\]
### Step 3: Substitute \( \tan x \)
Let \( T = \tan x \). Then, \( dx = \frac{1}{\sec^2 x} \, dT \) or \( dx = \frac{1}{1 + T^2} \, dT \). Also, we have:
\[
\sin x = \frac{T}{\sqrt{1 + T^2}}, \quad \cos x = \frac{1}{\sqrt{1 + T^2}}
\]
Thus, \( \sin^3 x = \left(\frac{T}{\sqrt{1 + T^2}}\right)^3 = \frac{T^3}{(1 + T^2)^{3/2}} \) and \( \cos x = \frac{1}{\sqrt{1 + T^2}} \).
### Step 4: Substitute into the Integral
Now substituting these into the integral:
\[
\int \frac{1}{\sqrt{\frac{T^3}{(1 + T^2)^{3/2}} \cdot \frac{1}{\sqrt{1 + T^2}}}} \cdot \frac{1}{1 + T^2} \, dT
\]
This simplifies to:
\[
\int \frac{(1 + T^2)^{2}}{\sqrt{T^3}} \cdot \frac{1}{1 + T^2} \, dT
\]
This reduces to:
\[
\int \frac{(1 + T^2)^{3/2}}{T^{3/2}} \, dT
\]
### Step 5: Integrate
Now we can integrate:
\[
\int T^{-3/2} (1 + T^2)^{3/2} \, dT
\]
Using the integration formula:
\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C
\]
we can solve this integral.
### Step 6: Back Substitute
After integrating, we substitute back \( T = \tan x \):
\[
\text{Final Result} = -\frac{2}{\sqrt{\tan x}} + C
\]
### Final Answer
Thus, the integral evaluates to:
\[
-\frac{2}{\sqrt{\tan x}} + C
\]
