To evaluate the integral \( I = \int \sqrt{\cot x} \, dx \), we can follow these steps:
### Step 1: Substitution
Let \( t = \cot x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[
dt = -\csc^2 x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\csc^2 x}
\]
Using the identity \( \csc^2 x = 1 + \cot^2 x = 1 + t^2 \), we can rewrite \( dx \):
\[
dx = -\frac{dt}{1 + t^2}
\]
### Step 2: Rewrite the Integral
Now substitute \( t \) into the integral:
\[
I = \int \sqrt{t} \left(-\frac{dt}{1 + t^2}\right) = -\int \frac{\sqrt{t}}{1 + t^2} \, dt
\]
### Step 3: Split the Integral
We can split the integral into two parts:
\[
I = -\int \frac{t^{1/2}}{1 + t^2} \, dt
\]
To evaluate this integral, we can use the method of partial fractions or substitution.
### Step 4: Use Partial Fractions
We can express \( \frac{t^{1/2}}{1 + t^2} \) in a more manageable form. However, it is easier to use the substitution \( u = t^{1/2} \), which gives \( t = u^2 \) and \( dt = 2u \, du \):
\[
I = -\int \frac{u}{1 + u^4} \cdot 2u \, du = -2 \int \frac{u^2}{1 + u^4} \, du
\]
### Step 5: Further Simplification
Now we can simplify the integral:
\[
I = -2 \int \frac{u^2}{1 + u^4} \, du
\]
This integral can be solved using trigonometric substitution or recognizing it as a standard integral.
### Step 6: Solve the Integral
Using the formula for integrals of the form \( \int \frac{x^2}{1 + x^4} \, dx \), we can find:
\[
\int \frac{u^2}{1 + u^4} \, du = \frac{1}{2} \tan^{-1}(u^2) + C
\]
Thus,
\[
I = -2 \left( \frac{1}{2} \tan^{-1}(u^2) \right) + C = -\tan^{-1}(u^2) + C
\]
### Step 7: Back Substitute
Substituting back \( u = \sqrt{t} = \sqrt{\cot x} \):
\[
I = -\tan^{-1}(\cot x) + C
\]
### Step 8: Final Result
Using the identity \( \tan^{-1}(\cot x) = \frac{\pi}{2} - x \):
\[
I = -\left(\frac{\pi}{2} - x\right) + C = x - \frac{\pi}{2} + C
\]
Thus, the final result is:
\[
\int \sqrt{\cot x} \, dx = x - \frac{\pi}{2} + C
\]
