The value of the integral `int_(0)^(pi) (xdx)/(1+cos alpha sinx), 0 lt alpha lt pi`, is

Options

A. `(pi alpha)/(sin alpha)`
B. `(pi alpha)/(1+sin alpha)`
C. `(pi alpha)/(cos alpha)`
D. `(pi alpha)/(1+cos alpha)`

Answer

To solve the integral

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

we will use the property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

### Step 1: Substitute \( x \) with \( \pi - x \)

Let’s denote the integral as \( I \):

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

Now, we will substitute \( x \) with \( \pi - x \):

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

Since \( \sin(\pi - x) = \sin x \), we have:

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

### Step 2: Combine the two integrals

Now, we can add the two expressions for \( I \):

\[ 2I = \int_{0}^{\pi} \left( \frac{x \, dx}{1 + \cos \alpha \sin x} + \frac{\pi - x \, dx}{1 + \cos \alpha \sin x} \right). \]

This simplifies to:

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

### Step 3: Solve for \( I \)

We can now express \( I \) as:

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

### Step 4: Evaluate the integral

Next, we need to evaluate the integral:

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

To do this, we can use the substitution \( t = \tan \frac{x}{2} \), which gives:

\[ \sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2}. \]

Changing the limits, when \( x = 0 \), \( t = 0 \) and when \( x = \pi \), \( t \to \infty \):

\[ \int_{0}^{\pi} \frac{dx}{1 + \cos \alpha \sin x} = \int_{0}^{\infty} \frac{2 \, dt}{(1 + t^2)(1 + \cos \alpha \cdot \frac{2t}{1 + t^2})}. \]

### Step 5: Simplify the integral

This can be simplified further, and we can complete the square in the denominator. After some algebra, we will arrive at:

\[ \int_{0}^{\infty} \frac{dt}{t^2 + (1 - \cos^2 \alpha)} = \frac{1}{\sin \alpha} \cdot \frac{\pi}{2}. \]

### Final Step: Combine results

Thus, we have:

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

### Conclusion

The value of the integral is:

\[ \boxed{\frac{\pi^2}{2 \sin \alpha}}. \]