If `int(sqrt(cotx))/(sinxcosx)dx=Psqrt(cotx)+Q` , then P equals

Options

A. 1
B. 2
C. `-1`
D. `-2`

Explanation

To solve the integral \( I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} \, dx \) and find the value of \( P \) in the expression \( I = P \sqrt{\cot x} + Q \), we can 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 \( \cot x \) as \( \frac{\cos x}{\sin x} \), so:
\[ \sqrt{\cot x} = \sqrt{\frac{\cos x}{\sin x}} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \]
Thus, we can express the integral as:
\[ I = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x} \sin x \cos x} \, dx = \int \frac{\sqrt{\cos x}}{\sin^{3/2} x \cos x} \, dx \]

### Step 2: Simplify the integral
Next, we divide the numerator and denominator by \( \cos^2 x \):
\[ I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\cot x}}{\sin x} \cdot \frac{1}{\cos x} \, dx \]
This can be rewritten as:
\[ I = \int \sqrt{\cot x} \cdot \sec^2 x \, dx \]

### Step 3: Substitute \( t = \tan x \)
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). The integral becomes:
\[ I = \int \left(\frac{1}{\sqrt{t}}\right) \cdot dt \]
This simplifies to:
\[ I = \int t^{-3/2} \, dt \]

### Step 4: Integrate
The integral of \( t^{-3/2} \) is:
\[ \int t^{-3/2} \, dt = \frac{t^{-1/2}}{-1/2} + C = -2 t^{-1/2} + C \]
Substituting back \( t = \tan x \):
\[ I = -2 \sqrt{\tan x} + C \]

### Step 5: Express in terms of \( \sqrt{\cot x} \)
Using the identity \( \tan x = \frac{1}{\cot x} \):
\[ I = -2 \sqrt{\frac{1}{\cot x}} + C = -2 \frac{1}{\sqrt{\cot x}} + C \]
This can be rewritten as:
\[ I = -2 \sqrt{\cot x} + C \]

### Step 6: Identify \( P \)
Comparing with the expression \( I = P \sqrt{\cot x} + Q \), we identify:
\[ P = -2 \]

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