If `I_(1)=intsin^(-1)xdx and I_(2)sin^(-1)sqrt(-x^(2))dx,` then

Options

A. `I_(1)=I_(2)`
B. `I_(2):I_(1)=(pi)/(2)`
C. `I_(1)+I_(2)=(pi)/(2)`
D. `I_(1)+I_(2)=(pi)/(2)x`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and then analyze their relationship.

### Step-by-Step Solution:

1. **Define the Integrals**:
\[ I_1 = \int \sin^{-1}(x) \, dx \]
\[ I_2 = \int \sin^{-1}(\sqrt{1 - x^2}) \, dx \]

2. **Simplify \( I_2 \)**:
We know that:
\[ \sin^{-1}(\sqrt{1 - x^2}) = \cos^{-1}(x) \]
Therefore, we can rewrite \( I_2 \) as:
\[ I_2 = \int \cos^{-1}(x) \, dx \]

3. **Combine the Integrals**:
Now we can express \( I_1 + I_2 \):
\[ I_1 + I_2 = \int \sin^{-1}(x) \, dx + \int \cos^{-1}(x) \, dx \]
This can be combined into a single integral:
\[ I_1 + I_2 = \int \left( \sin^{-1}(x) + \cos^{-1}(x) \right) \, dx \]

4. **Use the Identity**:
We know from trigonometric identities that:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Therefore:
\[ I_1 + I_2 = \int \frac{\pi}{2} \, dx \]

5. **Integrate**:
Now we can integrate:
\[ I_1 + I_2 = \frac{\pi}{2} \int dx = \frac{\pi}{2} x + C \]
where \( C \) is the constant of integration.

6. **Conclusion**:
Thus, we have:
\[ I_1 + I_2 = \frac{\pi}{2} x + C \]

### Final Result:
The relationship between \( I_1 \) and \( I_2 \) is:
\[ I_1 + I_2 = \frac{\pi}{2} x + C \]