The maximum value of cos `(int_(2x)^(x^(2)) e^t sin^2 " t dt ")`

Options

A. `1/2`
B. 0
C. 1
D. non -existent

Explanation

To solve the problem of finding the maximum value of \( \cos\left(\int_{2x}^{x^2} e^t \sin^2 t \, dt\right) \), we will follow these steps:

### Step 1: Define the function
Let:
\[ f(x) = \cos\left(\int_{2x}^{x^2} e^t \sin^2 t \, dt\right) \]

### Step 2: Identify the maximum value of cosine
The maximum value of the cosine function, \( \cos(\theta) \), is 1, which occurs when \( \theta = 0 \) (or any integer multiple of \( 2\pi \)). Therefore, to find the maximum value of \( f(x) \), we need to find when:
\[ \int_{2x}^{x^2} e^t \sin^2 t \, dt = 0 \]

### Step 3: Analyze the integral
The integral \( \int_{2x}^{x^2} e^t \sin^2 t \, dt \) will be equal to zero if the limits of integration are the same, or if the integrand evaluates to zero over the interval. The simplest case is when the limits are equal:
\[ 2x = x^2 \]

### Step 4: Solve the equation
Rearranging the equation \( 2x = x^2 \):
\[ x^2 - 2x = 0 \]
Factoring gives:
\[ x(x - 2) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x = 2 \]

### Step 5: Evaluate the function at the critical points
Now, we evaluate \( f(x) \) at the critical points:
1. For \( x = 0 \):
\[ f(0) = \cos\left(\int_{0}^{0} e^t \sin^2 t \, dt\right) = \cos(0) = 1 \]

2. For \( x = 2 \):
\[ f(2) = \cos\left(\int_{4}^{4} e^t \sin^2 t \, dt\right) = \cos(0) = 1 \]

### Step 6: Conclusion
Since both critical points yield \( f(x) = 1 \), the maximum value of \( f(x) \) is:
\[ \boxed{1} \]