The value of
`overset(sin^(2)x)underset(0)int sin^(-1)sqrt(t)dt+overset(cos^(2)x)underset(0)int cos^(-1)sqrt(t)dt`, is

Answer

To solve the problem, we need to evaluate the expression:

\[ I = \int_0^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt + \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt \]

### Step 1: Define the Integrals
Let:
\[ I_1 = \int_0^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt \]
\[ I_2 = \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt \]
Then, we have:
\[ I = I_1 + I_2 \]

### Step 2: Evaluate \(I_1\)
For \(I_1\), we will use the substitution \(t = \sin^2 \theta\). Then, we have:
\[ dt = 2\sin \theta \cos \theta \, d\theta = \sin(2\theta) \, d\theta \]
The limits change as follows:
- When \(t = 0\), \(\theta = 0\)
- When \(t = \sin^2 x\), \(\theta = x\)

Thus, we can rewrite \(I_1\):
\[ I_1 = \int_0^x \sin^{-1}(\sqrt{\sin^2 \theta}) \sin(2\theta) \, d\theta \]
Since \(\sin^{-1}(\sin \theta) = \theta\):
\[ I_1 = \int_0^x \theta \sin(2\theta) \, d\theta \]

### Step 3: Evaluate \(I_2\)
For \(I_2\), we use the substitution \(t = \cos^2 \theta\):
\[ dt = -2\cos \theta \sin \theta \, d\theta = -\sin(2\theta) \, d\theta \]
The limits change as follows:
- When \(t = 0\), \(\theta = \frac{\pi}{2}\)
- When \(t = \cos^2 x\), \(\theta = x\)

Thus, we can rewrite \(I_2\):
\[ I_2 = \int_{\frac{\pi}{2}}^x \cos^{-1}(\sqrt{\cos^2 \theta}) (-\sin(2\theta)) \, d\theta \]
Since \(\cos^{-1}(\cos \theta) = \theta\):
\[ I_2 = \int_x^{\frac{\pi}{2}} \theta \sin(2\theta) \, d\theta \]

### Step 4: Combine \(I_1\) and \(I_2\)
Now, we combine \(I_1\) and \(I_2\):
\[ I = \int_0^x \theta \sin(2\theta) \, d\theta + \int_x^{\frac{\pi}{2}} \theta \sin(2\theta) \, d\theta \]
Using the property of integrals:
\[ I = \int_0^{\frac{\pi}{2}} \theta \sin(2\theta) \, d\theta \]

### Step 5: Evaluate the Final Integral
Now we need to evaluate:
\[ \int_0^{\frac{\pi}{2}} \theta \sin(2\theta) \, d\theta \]
Using integration by parts, let:
- \(u = \theta\) and \(dv = \sin(2\theta) d\theta\)
- Then, \(du = d\theta\) and \(v = -\frac{1}{2} \cos(2\theta)\)

Applying integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
\[ = -\frac{1}{2} \theta \cos(2\theta) \bigg|_0^{\frac{\pi}{2}} + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos(2\theta) d\theta \]
Calculating the boundary term:
\[ = -\frac{1}{2} \left( \frac{\pi}{2} \cdot \cos(\pi) - 0 \cdot \cos(0) \right) + \frac{1}{2} \left( \frac{\sin(2\theta)}{2} \bigg|_0^{\frac{\pi}{2}} \right) \]
\[ = -\frac{1}{2} \left( \frac{\pi}{2} \cdot (-1) \right) + 0 \]
\[ = \frac{\pi}{4} \]

### Final Answer
Thus, the value of the original expression is:
\[ \boxed{\frac{\pi}{4}} \]