If `int_(0)^(1) cot^(-1)(1-x+x^(2))dx=k int_(0)^(1) tan^(-1)x dx`, then k=`

Options

A. 1
B. 2
C. `pi`
D. `2pi`

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \) and express it in terms of \( \int_{0}^{1} \tan^{-1}(x) \, dx \).

### Step-by-step Solution:

1. **Define the integral:**
\[ I = \int_{0}^{1} \cot^{-1}(1 - x + x^2) \, dx \]

2. **Use the identity for cotangent inverse:**
Recall that \( \cot^{-1}(y) = \tan^{-1}\left(\frac{1}{y}\right) \). Thus, we can rewrite the integral:
\[ I = \int_{0}^{1} \tan^{-1}\left(\frac{1}{1 - x + x^2}\right) \, dx \]

3. **Simplify the argument of the tangent inverse:**
The expression \( 1 - x + x^2 \) can be rewritten as:
\[ 1 - x + x^2 = (x - 1)^2 + 1 \]
Therefore,
\[ \frac{1}{1 - x + x^2} = \frac{1}{(x - 1)^2 + 1} \]

4. **Apply the tangent inverse identity:**
We can use the formula:
\[ \tan^{-1}(a - b) = \tan^{-1}(a) - \tan^{-1}(b) \text{ if } ab < 1 \]
Here, we can express:
\[ I = \int_{0}^{1} \left( \tan^{-1}(x) - \tan^{-1}(1 - x) \right) \, dx \]

5. **Evaluate the second integral:**
Let \( I_1 = \int_{0}^{1} \tan^{-1}(1 - x) \, dx \). We can use the property of definite integrals:
\[ I_1 = \int_{0}^{1} \tan^{-1}(1 - x) \, dx = \int_{0}^{1} \tan^{-1}(x) \, dx \]
This means:
\[ I_1 = \int_{0}^{1} \tan^{-1}(x) \, dx \]

6. **Combine the integrals:**
Therefore, we have:
\[ I = \int_{0}^{1} \tan^{-1}(x) \, dx - \int_{0}^{1} \tan^{-1}(1 - x) \, dx = \int_{0}^{1} \tan^{-1}(x) \, dx - I_1 \]
Since \( I_1 = I \), we can write:
\[ I = \int_{0}^{1} \tan^{-1}(x) \, dx - I \]
This implies:
\[ 2I = \int_{0}^{1} \tan^{-1}(x) \, dx \]
Hence,
\[ I = \frac{1}{2} \int_{0}^{1} \tan^{-1}(x) \, dx \]

7. **Relate it to the original equation:**
From the problem statement, we have:
\[ I = k \int_{0}^{1} \tan^{-1}(x) \, dx \]
Setting the two expressions for \( I \) equal gives:
\[ \frac{1}{2} \int_{0}^{1} \tan^{-1}(x) \, dx = k \int_{0}^{1} \tan^{-1}(x) \, dx \]
Dividing both sides by \( \int_{0}^{1} \tan^{-1}(x) \, dx \) (which is non-zero) yields:
\[ k = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( k \) is \( 2 \).