Let `g(t) = int_(x_(1))^(x^(2))f(t,x) dx`. Then `g'(t) = int_(x_(1))^(x^(2))(del)/(delt) (f(t,x))dx`, Consider `f(x) = int_(0)^(pi) (ln(1+xcostheta))/(costheta) d theta`.
The number of critical point of `f(x)`, in the interior of its domain, is

Options

A. 0
B. 1
C. 2
D. infinitely many

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the necessary steps to find the number of critical points of the function \( f(x) \).

### Step 1: Define the function \( f(x) \)
Given:
\[ f(x) = \int_{0}^{\pi} \frac{\ln(1 + x \cos \theta)}{\cos \theta} \, d\theta \]

### Step 2: Differentiate \( f(x) \) with respect to \( x \)
Using Leibniz's rule for differentiation under the integral sign, we can differentiate \( f(x) \):
\[ f'(x) = \int_{0}^{\pi} \frac{\partial}{\partial x} \left( \frac{\ln(1 + x \cos \theta)}{\cos \theta} \right) d\theta \]
The derivative of \( \ln(1 + x \cos \theta) \) is:
\[ \frac{1}{1 + x \cos \theta} \cdot \cos \theta \]
Thus,
\[ f'(x) = \int_{0}^{\pi} \frac{\cos \theta}{(1 + x \cos \theta) \cos \theta} \, d\theta = \int_{0}^{\pi} \frac{1}{1 + x \cos \theta} \, d\theta \]

### Step 3: Solve for critical points
To find the critical points, we set \( f'(x) = 0 \):
\[ \int_{0}^{\pi} \frac{1}{1 + x \cos \theta} \, d\theta = 0 \]
However, the integral \( \int_{0}^{\pi} \frac{1}{1 + x \cos \theta} \, d\theta \) is always positive for \( x > -1 \) and \( x < 1 \). Therefore, we need to analyze the behavior of \( f'(x) \) to find critical points.

### Step 4: Analyze the function \( f'(x) \)
The function \( f'(x) \) is positive for \( x \) in the interval \( (-1, 1) \) and approaches infinity as \( x \) approaches -1 or 1. Since \( f'(x) \) does not equal zero in this interval, we can conclude that there are no critical points in the interior of the domain.

### Step 5: Conclusion
The number of critical points of \( f(x) \) in the interior of its domain is:
\[ \text{Number of critical points} = 0 \]