The value of `int sqrt((1-cosx)/(cosalpha-cosx)) dx` where `0ltalphaltxltpi` is equal to:

Options

A. `2"ln"(cosalpha/2-cosx/2)+C`
B. `sqrt(2)"ln"(cosalpha/2-cosx/2)+C`
C. `2sqrt(2)"ln"(cosalpha/2-cosx/2)+C`
D. `-2sin^(-1)(cosx/2)/(cosalpha/2)+C`

Answer

To solve the integral \( \int \sqrt{\frac{1 - \cos x}{\cos \alpha - \cos x}} \, dx \) where \( 0 < \alpha < \pi \) and \( 0 < x < \pi \), we will follow these steps:

### Step 1: Simplify the expression inside the integral
We know that:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ \int \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{\cos \alpha - \cos x}} \, dx \]

### Step 2: Rewrite \( \cos x \) using the half-angle formula
Using the half-angle formula for cosine:
\[ \cos x = 2 \cos^2\left(\frac{x}{2}\right) - 1 \]
We can express \( \cos \alpha - \cos x \) as:
\[ \cos \alpha - \cos x = \cos \alpha - (2 \cos^2\left(\frac{x}{2}\right) - 1) = \cos \alpha + 1 - 2 \cos^2\left(\frac{x}{2}\right) \]

### Step 3: Substitute into the integral
Now substituting back into the integral, we get:
\[ \int \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{\cos \alpha + 1 - 2 \cos^2\left(\frac{x}{2}\right)}} \, dx \]

### Step 4: Factor out constants
We can factor out constants from the square root:
\[ = \sqrt{2} \int \frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos \alpha + 1 - 2 \cos^2\left(\frac{x}{2}\right)}} \, dx \]

### Step 5: Use substitution
Let \( t = \cos\left(\frac{x}{2}\right) \), then \( dx = -\frac{2}{\sqrt{1 - t^2}} \, dt \). The limits of integration will change accordingly.

### Step 6: Change the limits of integration
When \( x = 0 \), \( t = 1 \) and when \( x = \pi \), \( t = 0 \). Thus, the integral becomes:
\[ -\sqrt{2} \int_{1}^{0} \frac{\sqrt{1 - t^2}}{\sqrt{\cos \alpha + 1 - 2t^2}} \cdot \frac{2}{\sqrt{1 - t^2}} \, dt \]
This simplifies to:
\[ 2\sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{\cos \alpha + 1 - 2t^2}} \, dt \]

### Step 7: Evaluate the integral
The integral can be evaluated using the standard result:
\[ \int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1}\left(\frac{u}{a}\right) + C \]
where \( a^2 = \cos \alpha + 1 \).

### Final Result
After evaluating the integral and substituting back, we find:
\[ = -2 \sqrt{2} \sin^{-1}\left(\frac{t}{\sqrt{\cos \alpha + 1}}\right) \bigg|_{0}^{1} \]
This leads us to the final answer.