Find the number points where `f (theta) = int _(-1)^(1) (sin theta dx )/(1-2x cos theta + x ^(2))` is discontinous where `thetain [0, 2pi].`

Answer

To find the number of points where the function

\[ f(\theta) = \int_{-1}^{1} \frac{\sin \theta}{1 - 2x \cos \theta + x^2} \, dx \]

is discontinuous for \(\theta \in [0, 2\pi]\), we follow these steps:

### Step 1: Analyze the integrand
The integrand is given by

\[ \frac{\sin \theta}{1 - 2x \cos \theta + x^2}. \]

We need to identify when this function becomes undefined or discontinuous. The denominator \(1 - 2x \cos \theta + x^2\) can be rewritten as:

\[ (x - \cos \theta)^2 + \sin^2 \theta. \]

### Step 2: Identify points of discontinuity
The function will be discontinuous when the denominator equals zero, which occurs when:

\[ (x - \cos \theta)^2 + \sin^2 \theta = 0. \]

This equation holds true if and only if both terms are zero:

1. \((x - \cos \theta)^2 = 0 \implies x = \cos \theta\)
2. \(\sin^2 \theta = 0 \implies \sin \theta = 0\)

### Step 3: Solve for \(\theta\)
From \(\sin \theta = 0\), we find the values of \(\theta\) within the interval \([0, 2\pi]\):

- \(\theta = 0\)
- \(\theta = \pi\)
- \(\theta = 2\pi\)

These are the points where \(\sin \theta\) is zero.

### Step 4: Check the corresponding \(x\) values
Now we check the corresponding \(x\) values:

- For \(\theta = 0\), \(\cos(0) = 1\) → \(x = 1\)
- For \(\theta = \pi\), \(\cos(\pi) = -1\) → \(x = -1\)
- For \(\theta = 2\pi\), \(\cos(2\pi) = 1\) → \(x = 1\)

### Step 5: Determine points of discontinuity
The function \(f(\theta)\) is discontinuous at:

- \(\theta = 0\) (where \(x = 1\))
- \(\theta = \pi\) (where \(x = -1\))
- \(\theta = 2\pi\) (where \(x = 1\))

However, since \(x = 1\) occurs for both \(\theta = 0\) and \(\theta = 2\pi\), we count it only once.

### Conclusion
Thus, the points of discontinuity for \(f(\theta)\) are:

- \(\theta = 0\)
- \(\theta = \pi\)
- \(\theta = 2\pi\)

This gives us a total of **3 points** of discontinuity.

### Final Answer
The number of points where \(f(\theta)\) is discontinuous is **3**.

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