If `I_(1)=int_(0)^(pi//2) cos(sin x) dx,I_(2)=int_(0)^(pi//2) sin (cos x) dx and I_(3)=int_(0)^(pi//2) cos x dx` then

Options

A. `I_(1)gtI_(3)gtI_(2)`
B. `I_(3)gtI_(1)gtI_(2)`
C. `I_(1)gtI_(2)gtI_(3)`
D. `I_(3)gtI_(2)gtI_(1)`

Explanation

To solve the problem, we need to analyze the three integrals \( I_1 \), \( I_2 \), and \( I_3 \):

1. **Define the Integrals**:
- \( I_1 = \int_{0}^{\frac{\pi}{2}} \cos(\sin x) \, dx \)
- \( I_2 = \int_{0}^{\frac{\pi}{2}} \sin(\cos x) \, dx \)
- \( I_3 = \int_{0}^{\frac{\pi}{2}} \cos x \, dx \)

2. **Evaluate \( I_3 \)**:
- The integral \( I_3 \) can be calculated directly:
\[ I_3 = \int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

3. **Compare \( I_1 \) and \( I_2 \)**:
- To compare \( I_1 \) and \( I_2 \), we will use the properties of the functions involved.
- For \( x \in [0, \frac{\pi}{2}] \):
- Since \( \sin x \) is increasing and \( \cos x \) is decreasing, we know:
- \( \sin x < x \) for \( x > 0 \)
- \( \cos x > \sin(\cos x) \) since \( \cos x \) is always positive in this interval.
- Thus, we can conclude that:
\[ \cos(\sin x) > \sin(\cos x) \]
- Therefore, integrating both sides from \( 0 \) to \( \frac{\pi}{2} \):
\[ I_1 > I_2 \]

4. **Combine the Results**:
- We have established:
- \( I_3 = 1 \)
- \( I_1 > I_2 \)
- To compare \( I_1 \) and \( I_3 \), we note that \( \cos(\sin x) \) is always greater than \( \cos x \) for \( x \in [0, \frac{\pi}{2}] \) since \( \sin x \) is always less than or equal to \( x \).
- Therefore, we can conclude:
\[ I_1 > I_3 \]

5. **Final Order**:
- From the comparisons, we have:
\[ I_1 > I_3 > I_2 \]

Thus, the final result is:
\[ I_1 > I_3 > I_2 \]