If `sum_(i=1)^(4)(sin^(-1)x_(i)+cos^(-1)y_(i))=6pi`, then `\int\limitsum_(i=1)^(4)x(i)^sum(i=1)^(4)y(i) xln(1+x^(2))(e^(x^(2))/(1+e^(2x)))dx` is euqal to

Options

A. 0
B. `e^(4)+e^(-4)`
C. `ln (17/12)`
D. `e^(4)-e^(-4)`

Answer

To solve the given problem, we will follow the steps outlined in the video transcript and break it down into a step-by-step solution.

### Step 1: Understand the Given Condition
We are given the equation:
\[ \sum_{i=1}^{4} (\sin^{-1} x_i + \cos^{-1} y_i) = 6\pi \]
From the properties of the inverse sine and cosine functions, we know:
- \(\sin^{-1} x\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\)
- \(\cos^{-1} y\) ranges from \(0\) to \(\pi\)

Given that the total is \(6\pi\), we can deduce that:
\[ \sin^{-1} x_i = \frac{\pi}{2} \quad \text{and} \quad \cos^{-1} y_i = \pi \]
for each \(i\).

### Step 2: Solve for \(x_i\) and \(y_i\)
From \(\sin^{-1} x_i = \frac{\pi}{2}\):
\[ x_i = \sin\left(\frac{\pi}{2}\right) = 1 \]
From \(\cos^{-1} y_i = \pi\):
\[ y_i = \cos(\pi) = -1 \]
Thus, we have:
\[ x_1 = x_2 = x_3 = x_4 = 1 \quad \text{and} \quad y_1 = y_2 = y_3 = y_4 = -1 \]

### Step 3: Calculate the Sums
Now we calculate the sums:
\[ \sum_{i=1}^{4} x_i = 4 \cdot 1 = 4 \]
\[ \sum_{i=1}^{4} y_i = 4 \cdot (-1) = -4 \]

### Step 4: Set Up the Integral
We need to evaluate the integral:
\[ \int \sum_{i=1}^{4} y_i \sum_{i=1}^{4} x_i \cdot x \ln(1+x^2) \cdot \frac{e^x}{1+e^{2x}} \, dx \]
Substituting the sums we found:
\[ \int (-4)(4) \cdot x \ln(1+x^2) \cdot \frac{e^x}{1+e^{2x}} \, dx = -16 \int x \ln(1+x^2) \cdot \frac{e^x}{1+e^{2x}} \, dx \]

### Step 5: Check if the Function is Even or Odd
To determine the value of the integral, we check if the integrand is even or odd. We replace \(x\) with \(-x\):
\[ f(-x) = -x \ln(1+(-x)^2) \cdot \frac{e^{-x}}{1+e^{-2x}} = -x \ln(1+x^2) \cdot \frac{1}{e^x(1+e^{-2x})} \]
This shows that:
\[ f(-x) = -f(x) \]
Thus, the function is odd.

### Step 6: Evaluate the Integral
Since the integral of an odd function over a symmetric interval around zero is zero:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
Therefore, we conclude:
\[ -16 \int x \ln(1+x^2) \cdot \frac{e^x}{1+e^{2x}} \, dx = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int \sum_{i=1}^{4} y_i \sum_{i=1}^{4} x_i \cdot x \ln(1+x^2) \cdot \frac{e^x}{1+e^{2x}} \, dx = 0 \]