`I=int_0^(2pi) e^(sin^2x+sinx+1)dx` then

Options

A. `pi e^(3) lt I lt 2pie^(5)`
B. `2pie^(3//4) I lt 2 pie^(3)`
C. `2pie^(3) lt I lt 2pie^(4)`
D. `0 lt I lt 2pi`

Answer

To solve the integral \( I = \int_0^{2\pi} e^{\sin^2 x + \sin x + 1} \, dx \), we can follow these steps:

### Step 1: Simplify the exponent
We start with the expression in the exponent:
\[ \sin^2 x + \sin x + 1 \]
We can rewrite this as:
\[ \sin^2 x + \sin x + 1 = \left( \sin x + \frac{1}{2} \right)^2 + 1 - \frac{1}{4} \]
This simplifies to:
\[ \left( \sin x + \frac{1}{2} \right)^2 + \frac{3}{4} \]

### Step 2: Substitute back into the integral
Now we substitute this back into the integral:
\[ I = \int_0^{2\pi} e^{\left( \sin x + \frac{1}{2} \right)^2 + \frac{3}{4}} \, dx \]
This can be factored as:
\[ I = e^{\frac{3}{4}} \int_0^{2\pi} e^{\left( \sin x + \frac{1}{2} \right)^2} \, dx \]

### Step 3: Analyze the function
Next, we need to analyze the function \( \left( \sin x + \frac{1}{2} \right)^2 \). The maximum value of \( \sin x \) is 1, so:
\[ \text{Maximum of } \left( \sin x + \frac{1}{2} \right)^2 = \left( 1 + \frac{1}{2} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \]
The minimum value occurs when \( \sin x = -1 \):
\[ \text{Minimum of } \left( \sin x + \frac{1}{2} \right)^2 = \left( -1 + \frac{1}{2} \right)^2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4} \]

### Step 4: Set up inequalities for the integral
Thus, we can establish the bounds for the integral:
\[ \frac{3}{4} \leq \left( \sin x + \frac{1}{2} \right)^2 \leq \frac{9}{4} \]
This implies:
\[ e^{\frac{3}{4}} \leq e^{\left( \sin x + \frac{1}{2} \right)^2} \leq e^{\frac{9}{4}} \]

### Step 5: Integrate the bounds
Now we can integrate these bounds over \( [0, 2\pi] \):
\[ \int_0^{2\pi} e^{\frac{3}{4}} \, dx \leq I \leq \int_0^{2\pi} e^{\frac{9}{4}} \, dx \]
Calculating these integrals gives:
\[ 2\pi e^{\frac{3}{4}} \leq I \leq 2\pi e^{\frac{9}{4}} \]

### Final Result
Thus, we can conclude:
\[ 2\pi e^{\frac{3}{4}} \leq I \leq 2\pi e^{\frac{9}{4}} \]