If `f(x)=p(x)q(x),` where `p(x)=sqrt(cosx),q(x)=log(((1-x))/(1+x))`, then `int_(a)^(b)f(x)dx` equals, where `a=-1//2,b=1//2`

Options

A. `2int_(0)^(1//2)p(x)q(x)dx`
B. `int_(0)^(b)p(x)q(x)`
C. 1
D. 0

Answer

To solve the integral \( \int_{-1/2}^{1/2} f(x) \, dx \) where \( f(x) = p(x) q(x) \), \( p(x) = \sqrt{\cos x} \) and \( q(x) = \log\left(\frac{1-x}{1+x}\right) \), we can follow these steps:

### Step 1: Analyze the Function
We start with the function:
\[ f(x) = \sqrt{\cos x} \cdot \log\left(\frac{1-x}{1+x}\right) \]

### Step 2: Check for Symmetry
We will check if \( f(x) \) is an odd function. An odd function satisfies the property \( f(-x) = -f(x) \).

Calculating \( f(-x) \):
\[ f(-x) = \sqrt{\cos(-x)} \cdot \log\left(\frac{1+x}{1-x}\right) \]
Since \( \cos(-x) = \cos x \), we have:
\[ f(-x) = \sqrt{\cos x} \cdot \log\left(\frac{1+x}{1-x}\right) \]

### Step 3: Simplify \( f(-x) \)
Using the property of logarithms:
\[ \log\left(\frac{1+x}{1-x}\right) = -\log\left(\frac{1-x}{1+x}\right) \]
Thus,
\[ f(-x) = \sqrt{\cos x} \cdot \left(-\log\left(\frac{1-x}{1+x}\right)\right) = -f(x) \]

### Step 4: Conclude that \( f(x) \) is Odd
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Evaluate the Integral
The integral of an odd function over a symmetric interval around zero is zero:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
In our case, \( a = \frac{1}{2} \):
\[ \int_{-1/2}^{1/2} f(x) \, dx = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-1/2}^{1/2} f(x) \, dx = 0 \]

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