If f and g are two continuous function, then the value of `int_(-pi//4)^(pi//4){f(x)+f(-x)}{g(x)-g(-x)}dx` is

Options

A. `pi/4`
B. `0`
C. `-pi/4`
D. `pi`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x) + f(-x)}{g(x) - g(-x)} \, dx, \]

we will use the properties of definite integrals and the symmetry of the functions involved.

### Step 1: Define the integral
Let

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x) + f(-x)}{g(x) - g(-x)} \, dx. \]

### Step 2: Substitute \(x\) with \(-x\)
Now, we will evaluate the integral by substituting \(x\) with \(-x\):

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(-x) + f(x)}{g(-x) - g(x)} \, (-dx). \]

Since \(dx\) changes to \(-dx\), we can rewrite the limits:

\[ I = \int_{\frac{\pi}{4}}^{-\frac{\pi}{4}} \frac{f(-x) + f(x)}{g(-x) - g(x)} \, dx = -\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(-x) + f(x)}{g(-x) - g(x)} \, dx. \]

### Step 3: Rearranging the integral
Notice that \(g(-x) - g(x) = -(g(x) - g(-x))\). Therefore, we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(-x) + f(x)}{-(g(x) - g(-x))} \, dx = -\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(-x) + f(x)}{g(x) - g(-x)} \, dx. \]

### Step 4: Combine the two expressions for \(I\)
Now, we have two expressions for \(I\):

1. \(I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x) + f(-x)}{g(x) - g(-x)} \, dx\)
2. \(I = -\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(-x) + f(x)}{g(x) - g(-x)} \, dx\)

Adding these two equations gives:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{f(x) + f(-x)}{g(x) - g(-x)} - \frac{f(x) + f(-x)}{g(x) - g(-x)} \right) \, dx = 0. \]

### Step 5: Conclusion
Thus, we find that:

\[ I = 0. \]

### Final Answer
The value of the integral is

\[ \boxed{0}. \]