`int_(0)^(2pi) e^(cos x) cosx (sin x) dx`=

Options

A. 1
B. `pi`
C. `2pi`
D. none

Answer

To solve the integral

\[ I = \int_{0}^{2\pi} e^{\cos x} \cos x \sin x \, dx, \]

we will use a substitution method. Let's break down the solution step by step.

### Step 1: Substitution
Let \( y = e^{\cos x} \). Then, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = e^{\cos x} (-\sin x) = -y \sin x. \]

This implies:

\[ dy = -y \sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -\frac{dy}{y}. \]

### Step 2: Change the limits of integration
When \( x = 0 \):

\[ y = e^{\cos(0)} = e^{1} = e. \]

When \( x = 2\pi \):

\[ y = e^{\cos(2\pi)} = e^{1} = e. \]

Thus, the limits of integration remain the same, from \( e \) to \( e \).

### Step 3: Rewrite the integral
Now we can rewrite the integral \( I \):

\[ I = \int_{e}^{e} y \cos x \left(-\frac{dy}{y}\right). \]

This simplifies to:

\[ I = -\int_{e}^{e} \cos x \, dy. \]

### Step 4: Evaluate the integral
Since the limits of integration are the same (from \( e \) to \( e \)), the integral evaluates to zero:

\[ I = 0. \]

### Final Result
Thus, the value of the integral is:

\[ \int_{0}^{2\pi} e^{\cos x} \cos x \sin x \, dx = 0. \]