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: Multiply and Divide by \(\sqrt{2}\)

We start by multiplying and dividing the numerator by \(\sqrt{2}\):

\[ I = \int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2}(e^{x - \frac{\pi}{4}} + 1)} \, dx. \]

### Step 2: Rewrite the Numerator

Notice that \(\frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right)\) and \(\frac{1}{\sqrt{2}} = \cos\left(\frac{\pi}{4}\right)\). Thus, we can rewrite the numerator:

\[ \sin x + \cos x = \sqrt{2}\left(\sin\left(\frac{\pi}{4}\right) \sin x + \cos\left(\frac{\pi}{4}\right) \cos x\right) = \sqrt{2} \cos\left(x - \frac{\pi}{4}\right). \]

So we have:

\[ I = \int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{\sqrt{2} \cos\left(x - \frac{\pi}{4}\right)}{e^{x - \frac{\pi}{4}} + 1} \, dx. \]

### Step 3: Change of Variables

Let \(t = x - \frac{\pi}{4}\). Then, \(dx = dt\), and we need to change the limits:

- When \(x = -\frac{3\pi}{4}\), \(t = -\frac{3\pi}{4} - \frac{\pi}{4} = -\pi\).
- When \(x = \frac{5\pi}{4}\), \(t = \frac{5\pi}{4} - \frac{\pi}{4} = \pi\).

Thus, the integral becomes:

\[ I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \cos t}{e^t + 1} \, dt. \]

### Step 4: Use the Property of Definite Integrals

Using the property of definite integrals, we know that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, we have:

\[ I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \cos(-t)}{e^{-t} + 1} \, dt = \int_{-\pi}^{\pi} \frac{\sqrt{2} \cos t}{e^{-t} + 1} \, dt. \]

### Step 5: Combine the Two Integrals

Now we have two expressions for \(I\):

1. \(I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \cos t}{e^t + 1} \, dt\) (Equation 1)
2. \(I = \int_{-\pi}^{\pi} \frac{\sqrt{2} \cos t}{e^{-t} + 1} \, dt\) (Equation 2)

Adding these two equations:

\[ 2I = \int_{-\pi}^{\pi} \left( \frac{\sqrt{2} \cos t}{e^t + 1} + \frac{\sqrt{2} \cos t}{e^{-t} + 1} \right) dt. \]

### Step 6: Simplify the Integral

The expression simplifies as follows:

\[ \frac{\sqrt{2} \cos t}{e^t + 1} + \frac{\sqrt{2} \cos t}{e^{-t} + 1} = \sqrt{2} \cos t \cdot \frac{(e^{-t} + 1) + (e^t + 1)}{(e^t + 1)(e^{-t} + 1)} = \sqrt{2} \cos t \cdot \frac{e^t + e^{-t} + 2}{(e^t + 1)(e^{-t} + 1)}. \]

### Step 7: Evaluate the Integral

This integral evaluates to zero because the integrand is an odd function over the symmetric interval \([- \pi, \pi]\):

\[ \int_{-\pi}^{\pi} \sin t \, dt = 0. \]

Thus, we conclude:

\[ 2I = 0 \implies I = 0. \]

### Final Answer

The value of the integral is

\[ \boxed{0}. \]