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

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. none of these

Explanation

To solve the given 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: Differentiate the Expression
To simplify the evaluation, we will differentiate \( I \) with respect to \( x \):

\[ \frac{dI}{dx} = \frac{d}{dx} \left( \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt \right) \]

Using Leibniz's rule for differentiation under the integral sign, we have:

\[ \frac{dI}{dx} = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) + \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) \]

### Step 2: Compute the Derivatives
Now, we compute the derivatives of \( \sin^2 x \) and \( \cos^2 x \):

\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x \]
\[ \frac{d}{dx}(\cos^2 x) = -2 \sin x \cos x \]

### Step 3: Substitute Back into the Derivative
Substituting these derivatives back into the expression for \( \frac{dI}{dx} \):

\[ \frac{dI}{dx} = \sin^{-1}(\sin x) \cdot (2 \sin x \cos x) + \cos^{-1}(\cos x) \cdot (-2 \sin x \cos x) \]

### Step 4: Simplify the Expression
Using the identities \( \sin^{-1}(\sin x) = x \) and \( \cos^{-1}(\cos x) = x \):

\[ \frac{dI}{dx} = x \cdot (2 \sin x \cos x) - x \cdot (2 \sin x \cos x) = 0 \]

### Step 5: Conclude the Result
Since \( \frac{dI}{dx} = 0 \), this implies that \( I \) is constant for all \( x \). To find the value of \( I \), we can evaluate it at a specific value of \( x \), say \( x = 0 \):

\[ I(0) = \int_{0}^{\sin^2(0)} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{\cos^2(0)} \cos^{-1}(\sqrt{t}) \, dt \]

Calculating these integrals:

\[ I(0) = \int_{0}^{0} \sin^{-1}(\sqrt{t}) \, dt + \int_{0}^{1} \cos^{-1}(\sqrt{t}) \, dt = 0 + \int_{0}^{1} \cos^{-1}(\sqrt{t}) \, dt \]

### Step 6: Evaluate the Remaining Integral
Using the substitution \( u = \sqrt{t} \) (thus \( t = u^2 \) and \( dt = 2u \, du \)):

\[ \int_{0}^{1} \cos^{-1}(\sqrt{t}) \, dt = \int_{0}^{1} \cos^{-1}(u) \cdot 2u \, du \]

This integral can be evaluated using integration by parts or known integral results. The result of this integral is \( \frac{\pi}{4} \).

### Final Result
Thus, we conclude that:

\[ I = \frac{\pi}{4} \]