If `int(sqrt(cotx))/(sinx cos x)dx=A sqrt(cotx)+B`, then A is equal to ______________

Answer

To solve the integral \( I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} \, dx \) and express it in the form \( A \sqrt{\cot x} + B \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} \, dx \]
We can rewrite the denominator \( \sin x \cos x \) as \( \frac{1}{2} \sin(2x) \):
\[ I = 2 \int \frac{\sqrt{\cot x}}{\sin(2x)} \, dx \]

### Step 2: Multiply and Divide by \(\csc^2 x\)
Next, we multiply and divide the integrand by \(\csc^2 x\):
\[ I = \int \frac{\sqrt{\cot x} \cdot \csc^2 x}{\sin x \cos x \cdot \csc^2 x} \, dx \]
This gives us:
\[ I = \int \frac{\sqrt{\cot x} \cdot \csc^2 x}{\cot x} \, dx \]
Since \(\cot x = \frac{\cos x}{\sin x}\), we can simplify:
\[ I = \int \frac{\sqrt{\cot x} \cdot \csc^2 x}{\cot x} \, dx = \int \frac{\csc^2 x}{\sqrt{\cot x}} \, dx \]

### Step 3: Substitute \( t = \cot x \)
Let \( t = \cot x \). Then, we have:
\[ dt = -\csc^2 x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\csc^2 x} \]
Substituting this into the integral gives:
\[ I = \int \frac{\csc^2 x}{\sqrt{t}} \left(-\frac{dt}{\csc^2 x}\right) = -\int \frac{1}{\sqrt{t}} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ -\int t^{-1/2} \, dt = -2 t^{1/2} + C = -2 \sqrt{\cot x} + C \]

### Step 5: Compare with the Given Form
We have:
\[ I = -2 \sqrt{\cot x} + C \]
Comparing this with the form \( A \sqrt{\cot x} + B \), we find that \( A = -2 \).

Thus, the value of \( A \) is:
\[ \boxed{-2} \]