To evaluate the integral
\[
I = \int_0^{\pi} \frac{x \, dx}{1 + \cos \alpha \sin x}
\]
where \(0 < \alpha < \pi\), we can follow these steps:
### Step 1: Set up the integral
We start with the integral:
\[
I = \int_0^{\pi} \frac{x \, dx}{1 + \cos \alpha \sin x}
\]
### Step 2: Use the property of definite integrals
We can use the property of definite integrals that states:
\[
\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx
\]
In our case, \(a = 0\) and \(b = \pi\), so we have:
\[
I = \int_0^{\pi} \frac{\pi - x \, dx}{1 + \cos \alpha \sin(\pi - x)}
\]
Since \(\sin(\pi - x) = \sin x\), we can rewrite the integral as:
\[
I = \int_0^{\pi} \frac{\pi - x \, dx}{1 + \cos \alpha \sin x}
\]
### Step 3: Combine the two integrals
Now we have two expressions for \(I\):
1. \(I = \int_0^{\pi} \frac{x \, dx}{1 + \cos \alpha \sin x}\)
2. \(I = \int_0^{\pi} \frac{\pi - x \, dx}{1 + \cos \alpha \sin x}\)
Adding these two equations gives:
\[
2I = \int_0^{\pi} \frac{x + (\pi - x) \, dx}{1 + \cos \alpha \sin x}
\]
This simplifies to:
\[
2I = \int_0^{\pi} \frac{\pi \, dx}{1 + \cos \alpha \sin x}
\]
### Step 4: Solve for \(I\)
Now we can solve for \(I\):
\[
I = \frac{1}{2} \int_0^{\pi} \frac{\pi \, dx}{1 + \cos \alpha \sin x}
\]
### Step 5: Evaluate the integral
Now we need to evaluate the integral:
\[
\int_0^{\pi} \frac{dx}{1 + \cos \alpha \sin x}
\]
To evaluate this integral, we can use the substitution \(t = \tan\left(\frac{x}{2}\right)\), which gives:
\[
\sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2}
\]
Changing the limits accordingly, when \(x = 0\), \(t = 0\) and when \(x = \pi\), \(t \to \infty\). Thus, we have:
\[
\int_0^{\pi} \frac{dx}{1 + \cos \alpha \sin x} = \int_0^{\infty} \frac{2 \, dt}{(1 + t^2)(1 + \cos \alpha \frac{2t}{1 + t^2})}
\]
This integral can be simplified further, but we can use a known result for this type of integral:
\[
\int_0^{\pi} \frac{dx}{1 + k \sin x} = \frac{\pi}{\sqrt{1 + k^2}} \quad \text{for } |k| < 1
\]
In our case, \(k = \cos \alpha\), so:
\[
\int_0^{\pi} \frac{dx}{1 + \cos \alpha \sin x} = \frac{\pi}{\sqrt{1 + \cos^2 \alpha}}
\]
### Step 6: Substitute back to find \(I\)
Now substituting this back into our expression for \(I\):
\[
I = \frac{1}{2} \cdot \pi \cdot \frac{\pi}{\sqrt{1 + \cos^2 \alpha}} = \frac{\pi^2}{2\sqrt{1 + \cos^2 \alpha}}
\]
### Final Answer
Thus, the value of the integral is:
\[
I = \frac{\pi^2}{2\sqrt{1 + \cos^2 \alpha}}
\]
