If `f(x)=(2-xcosx)/(2+xcosx)andg(x)= "log"_(e)x, (xgt0)` then the value of the integral `int_(-pi//4)^(pi//4)g(f(x))` dx is

Options

A. `log_(e )2`
B. `log_(e )e`
C. `log_(e )1`
D. `log_(e )3`

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} g(f(x)) \, dx \), where \( f(x) = \frac{2 - x \cos x}{2 + x \cos x} \) and \( g(x) = \log_e x \), we can follow these steps:

### Step 1: Define the function \( g(f(x)) \)
We start by substituting \( f(x) \) into \( g(x) \):
\[ g(f(x)) = \log_e\left(\frac{2 - x \cos x}{2 + x \cos x}\right) \]

### Step 2: Evaluate \( g(f(-x)) \)
Next, we need to evaluate \( g(f(-x)) \):
\[ f(-x) = \frac{2 - (-x) \cos(-x)}{2 + (-x) \cos(-x)} = \frac{2 + x \cos x}{2 - x \cos x} \]
Thus,
\[ g(f(-x)) = \log_e\left(f(-x)\right) = \log_e\left(\frac{2 + x \cos x}{2 - x \cos x}\right) \]

### Step 3: Relate \( g(f(x)) \) and \( g(f(-x)) \)
Now, we can express \( g(f(-x)) \) in terms of \( g(f(x)) \):
\[ g(f(-x)) = \log_e\left(\frac{2 + x \cos x}{2 - x \cos x}\right) = -\log_e\left(\frac{2 - x \cos x}{2 + x \cos x}\right) = -g(f(x)) \]

### Step 4: Show that \( g(f(x)) \) is an odd function
From the previous step, we see that:
\[ g(f(-x)) = -g(f(x)) \]
This indicates that \( g(f(x)) \) is an odd function.

### Step 5: Use the property of odd functions in integrals
Since \( g(f(x)) \) is an odd function, we can use the property of integrals of odd functions:
\[ \int_{-a}^{a} h(x) \, dx = 0 \quad \text{for odd functions} \]
Thus,
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} g(f(x)) \, dx = 0 \]

### Conclusion
The value of the integral is:
\[ \boxed{0} \]