If `I_1 = int_t^1 1/(1 + x^2) dx and I_2 = int_1^(1//t) 1/(1 + x^2) dx` for `t gt 0` then

Options

A. `I_1 = I_2`
B. `I_1 gt I_2`
C. `I_2 gt I_1`
D. none of these

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and show that they are equal.

### Step-by-Step Solution:

1. **Define the Integrals**:
\[ I_1 = \int_t^1 \frac{1}{1 + x^2} \, dx \]
\[ I_2 = \int_1^{\frac{1}{t}} \frac{1}{1 + x^2} \, dx \]

2. **Evaluate \( I_1 \)**:
The integral \( \int \frac{1}{1 + x^2} \, dx \) is known to be \( \tan^{-1}(x) \). Therefore, we can evaluate \( I_1 \):
\[ I_1 = \left[ \tan^{-1}(x) \right]_t^1 = \tan^{-1}(1) - \tan^{-1}(t) \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ I_1 = \frac{\pi}{4} - \tan^{-1}(t) \]

3. **Evaluate \( I_2 \)**:
Similarly, we evaluate \( I_2 \):
\[ I_2 = \left[ \tan^{-1}(x) \right]_1^{\frac{1}{t}} = \tan^{-1}\left(\frac{1}{t}\right) - \tan^{-1}(1) \]
Again, substituting \( \tan^{-1}(1) = \frac{\pi}{4} \):
\[ I_2 = \tan^{-1}\left(\frac{1}{t}\right) - \frac{\pi}{4} \]

4. **Use the Identity for \( \tan^{-1} \)**:
We can use the identity:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Let \( a = 1 \) and \( b = t \):
\[ I_1 = \frac{\pi}{4} - \tan^{-1}(t) = \tan^{-1}\left(\frac{1 - t}{1 + t}\right) \]
And for \( I_2 \):
\[ I_2 = \tan^{-1}\left(\frac{1}{t}\right) - \frac{\pi}{4} = \tan^{-1}\left(\frac{\frac{1}{t} - 1}{1 + \frac{1}{t}}\right) \]

5. **Show \( I_1 = I_2 \)**:
Now we can see that:
\[ I_1 = \tan^{-1}\left(\frac{1 - t}{1 + t}\right) \]
\[ I_2 = \tan^{-1}\left(\frac{\frac{1 - t}{t}}{1 + \frac{1}{t}}\right) = \tan^{-1}\left(\frac{1 - t}{1 + t}\right) \]
Thus, we conclude that:
\[ I_1 = I_2 \]

### Final Result:
\[ I_1 = I_2 \]