If `I_n=int cot^n dc and I_0+I_1+2(I_2+.......+I_8)+I_9+I_10=A(u+(u^2)/2+........+(u^9)/9)+` constant where `u=cot x` then

Options

A. A=2
B. `A=-1`
C. A=1
D. A is dependent on x

Answer

To solve the problem, we need to evaluate the integral \( I_n = \int \cot^n x \, dx \) and find the expression for \( I_0 + I_1 + 2(I_2 + \ldots + I_8) + I_9 + I_{10} \).

### Step 1: Write the expression for \( I_n \)

We start with the integral:
\[ I_n = \int \cot^n x \, dx \]

### Step 2: Use integration by parts or reduction formula

We can use the reduction formula for integrals of powers of cotangent:
\[ I_n = \int \cot^n x \, dx = \frac{1}{n} \cot^{n-1} x - \frac{n-1}{n} I_{n-2} \]
This means we can express \( I_n \) in terms of \( I_{n-2} \).

### Step 3: Set up the equation

From the reduction formula, we can derive:
\[ I_n + I_{n-2} = \frac{1}{n-1} \cot^{n-1} x \]

### Step 4: Calculate the sum \( I_0 + I_1 + 2(I_2 + \ldots + I_8) + I_9 + I_{10} \)

We need to calculate:
\[ S = I_0 + I_1 + 2(I_2 + I_3 + I_4 + I_5 + I_6 + I_7 + I_8) + I_9 + I_{10} \]

### Step 5: Substitute values for \( I_n \)

Using the reduction formula, we can compute the values of \( I_n \) for \( n = 0 \) to \( n = 10 \):
- \( I_0 = \int dx = x + C \)
- \( I_1 = \int \cot x \, dx = \ln |\sin x| + C \)
- \( I_2 = \int \cot^2 x \, dx = -\cot x + x + C \)
- Continuing this process, we find the values for \( I_3, I_4, \ldots, I_{10} \).

### Step 6: Combine the results

After computing \( I_n \) for \( n = 0 \) to \( n = 10 \), we substitute these values into the expression for \( S \):
\[ S = I_0 + I_1 + 2(I_2 + I_3 + I_4 + I_5 + I_6 + I_7 + I_8) + I_9 + I_{10} \]

### Step 7: Substitute \( u = \cot x \)

Since \( u = \cot x \), we can express \( S \) in terms of \( u \):
\[ S = A\left(u + \frac{u^2}{2} + \ldots + \frac{u^9}{9}\right) + \text{constant} \]

### Step 8: Identify the value of \( A \)

By comparing the coefficients from the expression derived from the integral with the given expression, we can determine the value of \( A \).

### Conclusion

After evaluating the integrals and substituting, we find that:
\[ A = -1 \]