Evaluate `int_(0)^(pi)(x dx)/(1+cos alpha sin x)`,where `0lt alpha lt pi`.

Explanation

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 as given:

\[ I = \int_{0}^{\pi} \frac{x \, dx}{1 + \cos \alpha \sin x} \]

### 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 \( \sin(\pi - x) = \sin x \), this simplifies to:

\[ I = \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \cos \alpha \sin x} \]

### Step 3: Rewrite the integral
Now, we can split the integral into two parts:

\[ I = \int_{0}^{\pi} \frac{\pi \, dx}{1 + \cos \alpha \sin x} - \int_{0}^{\pi} \frac{x \, dx}{1 + \cos \alpha \sin x} \]

Notice that the second term is our original integral \( I \):

\[ I = \frac{\pi}{1 + \cos \alpha \sin x} - I \]

### Step 4: Solve for \( I \)
Now, we can add \( I \) to both sides:

\[ 2I = \int_{0}^{\pi} \frac{\pi \, dx}{1 + \cos \alpha \sin x} \]

Thus, we have:

\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi \, dx}{1 + \cos \alpha \sin x} \]

### Step 5: Evaluate the integral
Now we need to evaluate the integral:

\[ \int_{0}^{\pi} \frac{dx}{1 + \cos \alpha \sin x} \]

Using the substitution \( t = \tan\left(\frac{x}{2}\right) \), we have \( \sin x = \frac{2t}{1+t^2} \) and \( dx = \frac{2 \, dt}{1+t^2} \). The limits change from \( 0 \) to \( \infty \) as \( x \) goes from \( 0 \) to \( \pi \).

Substituting these into the integral gives:

\[ \int_{0}^{\infty} \frac{2 \, dt}{(1 + \cos \alpha \cdot \frac{2t}{1+t^2})(1+t^2)} \]

This integral can be evaluated using standard techniques, leading to:

\[ \int_{0}^{\pi} \frac{dx}{1 + \cos \alpha \sin x} = \frac{\pi}{\sin \alpha} \]

### Step 6: Substitute back to find \( I \)
Now substituting this back into our equation for \( I \):

\[ I = \frac{1}{2} \cdot \frac{\pi^2}{\sin \alpha} \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{\pi^2}{2 \sin \alpha} \]