To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cot x}{\cot x + \csc x} \, dx \) and express it in the form \( m\pi + n \), we can follow these steps:
### Step 1: Rewrite the integrand
We start with the integrand:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\cot x}{\cot x + \csc x} \, dx
\]
We know that:
\[
\cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \csc x = \frac{1}{\sin x}
\]
Thus, we can rewrite the integrand:
\[
\cot x + \csc x = \frac{\cos x}{\sin x} + \frac{1}{\sin x} = \frac{\cos x + 1}{\sin x}
\]
So, we have:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\cot x}{\frac{\cos x + 1}{\sin x}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\cot x \cdot \sin x}{\cos x + 1} \, dx
\]
This simplifies to:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + 1} \, dx
\]
### Step 2: Use symmetry of the integral
Now, we can use the property of definite integrals:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x + 1} \, dx
\]
We can also consider:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \csc x} \, dx
\]
This gives us:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \frac{1}{\sin x}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin^2 x + 1} \, dx
\]
### Step 3: Combine the two integrals
Now, we can add the two forms of \( I \):
\[
2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{\cos x}{\cos x + 1} + \frac{\sin x}{\sin x + 1} \right) \, dx
\]
This simplifies to:
\[
2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}
\]
Thus, we find:
\[
I = \frac{\pi}{4}
\]
### Step 4: Express in the form \( m\pi + n \)
Now we can express \( I \) in the form \( m\pi + n \):
\[
I = \frac{\pi}{4} = \frac{1}{4}\pi + 0
\]
Here, \( m = \frac{1}{4} \) and \( n = 0 \).
### Step 5: Calculate \( m \cdot n \)
Now, we compute:
\[
m \cdot n = \frac{1}{4} \cdot 0 = 0
\]
### Final Answer
Thus, the value of \( m \cdot n \) is \( 0 \).
