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 compare the values of the integrals \(I_1\), \(I_2\), and \(I_3\) defined as follows:

\[ I_1 = \int_{0}^{\frac{\pi}{2}} \cos(\sin x) \, dx \]
\[ I_2 = \int_{0}^{\frac{\pi}{2}} \sin(\cos x) \, dx \]
\[ I_3 = \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]

### Step 1: Evaluate \(I_3\)

First, we calculate \(I_3\):

\[ I_3 = \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]

The integral of \(\cos x\) is \(\sin x\). Thus,

\[ I_3 = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 2: Compare \(I_1\) and \(I_2\)

Next, we need to compare \(I_1\) and \(I_2\). To do this, we can use the properties of the functions involved.

1. **For \(I_1\)**:
- Since \(\sin x\) is increasing on \([0, \frac{\pi}{2}]\), \(\cos(\sin x)\) is a decreasing function because \(\cos\) is decreasing in the range of \(\sin x\) from \(0\) to \(1\).
- Therefore, \(\cos(\sin x) \geq \cos(1)\) for \(x \in [0, \frac{\pi}{2}]\).

2. **For \(I_2\)**:
- Since \(\cos x\) is decreasing on \([0, \frac{\pi}{2}]\), \(\sin(\cos x)\) is increasing because \(\sin\) is increasing in the range of \(\cos x\) from \(1\) to \(0\).
- Therefore, \(\sin(\cos x) \leq \sin(1)\) for \(x \in [0, \frac{\pi}{2}]\).

### Step 3: Establishing the Inequalities

Now we can establish the inequalities:

- Since \(\cos(\sin x) \geq \cos(1)\) and \(\sin(\cos x) \leq \sin(1)\), we can conclude that:

\[ I_1 = \int_{0}^{\frac{\pi}{2}} \cos(\sin x) \, dx \geq \int_{0}^{\frac{\pi}{2}} \sin(1) \, dx \]

- Since \(\sin(1) < 1\), we can also say that:

\[ I_2 = \int_{0}^{\frac{\pi}{2}} \sin(\cos x) \, dx \leq \int_{0}^{\frac{\pi}{2}} \sin(1) \, dx \]

### Step 4: Conclusion

From the evaluations and comparisons, we have:

\[ I_3 = 1 > I_2 \]
And since \(I_1\) is bounded below by \(\cos(1)\) and is greater than \(I_2\), we can conclude:

\[ I_1 > I_2 \]

Thus, we have:

\[ I_3 > I_2 \quad \text{and} \quad I_1 > I_2 \]

### Final Result

The order of the integrals is:

\[ I_3 > I_1 > I_2 \]