Let `I=int_(0)^(1) (sin x)/(sqrt(x))dx` and `J=int_(0)^(1) (cos x)/(sqrt(x)) dx`.
Then , which one of the following is true?

Options

A. `I gt (2)/(3)" and "J gt 2`
B. `I lt (2)/(3)" and "J lt 2`
C. `I lt (2)/(3)" and "J gt 2`
D. `I gt (2)/(3)" and "J lt 2`

Explanation

To solve the problem, we need to evaluate the integrals \( I \) and \( J \) defined as follows:

\[ I = \int_0^1 \frac{\sin x}{\sqrt{x}} \, dx \]
\[ J = \int_0^1 \frac{\cos x}{\sqrt{x}} \, dx \]

### Step 1: Estimate \( I \)

We know that for \( x \in [0, 1] \), \( \sin x \) is less than or equal to \( x \). Therefore, we can write:

\[ \frac{\sin x}{\sqrt{x}} \leq \frac{x}{\sqrt{x}} = \sqrt{x} \]

Now, we can integrate both sides from 0 to 1:

\[ I = \int_0^1 \frac{\sin x}{\sqrt{x}} \, dx \leq \int_0^1 \sqrt{x} \, dx \]

Calculating the right-hand side:

\[ \int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \left( 1^{3/2} - 0^{3/2} \right) = \frac{2}{3} \]

Thus, we have:

\[ I < \frac{2}{3} \]

### Step 2: Estimate \( J \)

Next, we consider \( J \). For \( x \in [0, 1] \), \( \cos x \) is less than or equal to 1. Therefore, we can write:

\[ \frac{\cos x}{\sqrt{x}} \leq \frac{1}{\sqrt{x}} \]

Integrating both sides from 0 to 1 gives:

\[ J = \int_0^1 \frac{\cos x}{\sqrt{x}} \, dx \leq \int_0^1 \frac{1}{\sqrt{x}} \, dx \]

Calculating the right-hand side:

\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \int_0^1 x^{-1/2} \, dx = \left[ \frac{x^{1/2}}{1/2} \right]_0^1 = 2 \left( 1^{1/2} - 0^{1/2} \right) = 2 \]

Thus, we have:

\[ J < 2 \]

### Conclusion

From our estimates, we have:

\[ I < \frac{2}{3} \quad \text{and} \quad J < 2 \]

This implies that \( I < J \).

### Final Answer

Thus, the correct relationship between \( I \) and \( J \) is:

\[ I < J \]