`int(dx)/(sqrt(sin^(3)xcosx))=?`

Options

A. `(-2)/(sqrt(tanx))+C`
B. `2sqrt(tanx)+C`
C. `(2)/(sqrt(tanx))+C`
D. `-2sqrt(tanx)+C`

Answer

To solve the integral \( I = \int \frac{dx}{\sqrt{\sin^3 x \cos x}} \), we will follow a systematic approach.

### Step 1: Rewrite the integral
We can express the integral as:
\[ I = \int \frac{dx}{\sqrt{\sin^3 x} \sqrt{\cos x}} \]

### Step 2: Multiply and divide by \( \cos^3 x \)
To simplify the integral, we multiply and divide by \( \cos^3 x \):
\[ I = \int \frac{\cos^3 x \, dx}{\sqrt{\sin^3 x \cos^4 x}} \]

### Step 3: Simplify the expression under the square root
Now we can rewrite the integral:
\[ I = \int \frac{\cos^3 x \, dx}{\sqrt{\sin^3 x} \cdot \cos^2 x} \]
This simplifies to:
\[ I = \int \frac{\cos x \, dx}{\sqrt{\sin^3 x}} = \int \frac{dx}{\sqrt{\tan^3 x}} \]

### Step 4: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, \( \sec^2 x \, dx = dt \) or \( dx = \frac{dt}{\sec^2 x} \). Since \( \sec^2 x = 1 + \tan^2 x = 1 + t^2 \), we have:
\[ dx = \frac{dt}{1 + t^2} \]

### Step 5: Substitute in the integral
Substituting \( t \) into the integral gives:
\[ I = \int \frac{dt}{t^{3/2} (1 + t^2)} \]

### Step 6: Solve the integral
Using the integral formula for \( \int \frac{dt}{t^{3/2}(1 + t^2)} \), we can integrate:
\[ I = \int t^{-3/2} (1 + t^2)^{-1} dt \]
This integral can be solved using partial fractions or a trigonometric substitution.

### Step 7: Final result
After performing the integration and substituting back \( t = \tan x \), we find:
\[ I = -\frac{2}{\sqrt{\tan x}} + C \]

### Conclusion
Thus, the final answer is:
\[ I = -\frac{2}{\sqrt{\tan x}} + C \]