`int_(0)^(pi//2) cosx\ e^(sinx)\ dx` is equal to

Options

A. `e+1`
B. `e-1`
C. `e`
D. `-e`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \cos x \, e^{\sin x} \, dx \), we will use a substitution method. Here are the steps:

### Step 1: Set up the substitution
Let \( u = \sin x \). Then, we differentiate to find \( du \):
\[ du = \cos x \, dx \]
This means that \( \cos x \, dx = du \).

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ u = \sin(0) = 0 \]
When \( x = \frac{\pi}{2} \):
\[ u = \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus, the limits of integration change from \( x: 0 \to \frac{\pi}{2} \) to \( u: 0 \to 1 \).

### Step 3: Rewrite the integral in terms of \( u \)
Now we can rewrite the integral:
\[ I = \int_{0}^{1} e^{u} \, du \]

### Step 4: Integrate
The integral of \( e^{u} \) is simply \( e^{u} \):
\[ I = \left[ e^{u} \right]_{0}^{1} = e^{1} - e^{0} \]
Calculating this gives:
\[ I = e - 1 \]

### Final Answer
Thus, the value of the integral is:
\[ I = e - 1 \]