The value of the integral `int_(-3pi//4)^(5pi//4)((sinx+cosx))/(e^(x-pi//4)+1)dx` is

Options

A. `0`
B. `1`
C. `2`
D. none of these

Explanation

To solve the integral

\[ I = \int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{\sin x + \cos x}{e^{(x - \frac{\pi}{4})} + 1} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integral
Let \( I = \int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{\sin x + \cos x}{e^{(x - \frac{\pi}{4})} + 1} \, dx \).

### Step 2: Use Trigonometric Identities
We can express \( \sin x + \cos x \) in terms of a single sine function:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right). \]
Thus, we rewrite the integral:
\[ I = \int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}{e^{(x - \frac{\pi}{4})} + 1} \, dx. \]

### Step 3: Change of Variables
Now, we perform a substitution \( u = x - \frac{\pi}{4} \), which gives \( du = dx \). The limits change as follows:
- When \( x = -\frac{3\pi}{4} \), \( u = -\frac{3\pi}{4} - \frac{\pi}{4} = -\pi \).
- When \( x = \frac{5\pi}{4} \), \( u = \frac{5\pi}{4} - \frac{\pi}{4} = \pi \).

Thus, the integral becomes:
\[ I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \sin\left(u + \frac{\pi}{4}\right)}{e^u + 1} \, du. \]

### Step 4: Symmetry of the Integral
Now, we can analyze the integral:
\[ I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \sin\left(u + \frac{\pi}{4}\right)}{e^u + 1} \, du. \]
Notice that \( \sin\left(u + \frac{\pi}{4}\right) \) is an odd function, and \( e^u + 1 \) is an even function. The product of an odd function and an even function is odd.

### Step 5: Evaluate the Integral
Since the integral of an odd function over a symmetric interval around zero is zero, we conclude:
\[ I = 0. \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{0}. \]