$\int_{-\frac{1}{2}}^{\frac{1}{2}}cos~x~log_{e}(\frac{1+x}{1-x})dx=$

Q: $\int_{-\frac{1}{2}}^{\frac{1}{2}}cos~x~log_{e}(\frac{1+x}{1-x})dx=$

To solve the integral $\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos x \log_e\left(\frac{1+x}{1-x}\right) dx$, we need to analyze the symmetry and properties of the integrand. ### Explanation for Option 2 (Correct Answer: 0) 1. **Symmetry of the Integrand:** - The function $\cos x$ is an even function, meaning $\cos(-x) = \cos x$. - The function $\log_e\left(\frac{1+x}{1-x}\right)$ is an odd function, meaning $\log_e\left(\frac{1-x}{1+x}\right) = -\log_e\left(\frac{1+x}{1-x}\right)$. 2. **Product of Even and Odd Functions:** - The product of an even function and an odd function is an odd function. Therefore, $\cos x \log_e\left(\frac{1+x}{1-x}\right)$ is an odd function. 3. **Integration of an Odd Function Over a Symmetric Interval:** - When integrating an odd function over a symmetric interval about the origin, the result is zero. This is because the contributions from the negative and positive halves of the interval cancel each other out. Given these points, the integral evaluates to zero, which supports Option 2 as the correct answer. ### Explanation for Why Other Options Are Incorrect: 1. **Option 1 (4):** - The integral of an odd function over a symmetric interval cannot be a non-zero value like 4. This option is incorrect because it contradicts the property of odd functions. 2. **Option 3 (2):** - Similar to Option 1, the integral of an odd function over a symmetric interval cannot yield a non-zero result like 2. This option is incorrect for the same reason. 3. **Option 4 ($\pi$):** - The value $\pi$ is unrelated to the integral of an odd function over a symmetric interval. This option is incorrect because it does not align with the mathematical properties of the integrand. In conclusion, the correct answer is Option 2 (0) due to the symmetry and properties of the integrand, which result in the integral evaluating to zero.

Question

Accepted Answer

To solve the integral $\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos x \log_e\left(\frac{1+x}{1-x}\right) dx$, we need to analyze the symmetry and properties of the integrand.

### Explanation for Option 2 (Correct Answer: 0)

1. **Symmetry of the Integrand:**
   - The function $\cos x$ is an even function, meaning $\cos(-x) = \cos x$.
   - The function $\log_e\left(\frac{1+x}{1-x}\right)$ is an odd function, meaning $\log_e\left(\frac{1-x}{1+x}\right) = -\log_e\left(\frac{1+x}{1-x}\right)$.

2. **Product of Even and Odd Functions:**
   - The product of an even function and an odd function is an odd function. Therefore, $\cos x \log_e\left(\frac{1+x}{1-x}\right)$ is an odd function.

3. **Integration of an Odd Function Over a Symmetric Interval:**
   - When integrating an odd function over a symmetric interval about the origin, the result is zero. This is because the contributions from the negative and positive halves of the interval cancel each other out.

Given these points, the integral evaluates to zero, which supports Option 2 as the correct answer.

### Explanation for Why Other Options Are Incorrect:

1. **Option 1 (4):**
   - The integral of an odd function over a symmetric interval cannot be a non-zero value like 4. This option is incorrect because it contradicts the property of odd functions.

2. **Option 3 (2):**
   - Similar to Option 1, the integral of an odd function over a symmetric interval cannot yield a non-zero result like 2. This option is incorrect for the same reason.

3. **Option 4 ($\pi$):**
   - The value $\pi$ is unrelated to the integral of an odd function over a symmetric interval. This option is incorrect because it does not align with the mathematical properties of the integrand.

In conclusion, the correct answer is Option 2 (0) due to the symmetry and properties of the integrand, which result in the integral evaluating to zero.