For $$0 \leq x \leq \frac{\pi}{2}$$, the value of $$\int_0^{\sin^2 x} \sin^{-1}(\sqrt{t})dt + \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t})dt$$ equals :

We need to evaluate the expression $$\int_0^{\sin^2 x} \sin^{-1}(\sqrt{t}) dt + \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t}) dt$$ for $$0 \leq x \leq \frac{\pi}{2}$$. Let's denote the first integral as $$I_1$$ and the second as $$I_2$$, so the expression is $$I = I_1 + I_2$$.

First, evaluate $$I_1 = \int_0^{\sin^2 x} \sin^{-1}(\sqrt{t}) dt$$. Use integration by parts, where $$u = \sin^{-1}(\sqrt{t})$$ and $$dv = dt$$. Then $$v = t$$, and we find $$du$$. Set $$w = \sqrt{t}$$, so $$u = \sin^{-1}(w)$$. Then $$\frac{du}{dw} = \frac{1}{\sqrt{1 - w^2}}$$, and $$\frac{dw}{dt} = \frac{1}{2\sqrt{t}}$$, so $$\frac{du}{dt} = \frac{1}{\sqrt{1 - w^2}} \cdot \frac{1}{2\sqrt{t}} = \frac{1}{\sqrt{1 - t}} \cdot \frac{1}{2\sqrt{t}}$$. Thus, $$du = \frac{1}{2\sqrt{t} \sqrt{1 - t}} dt$$.

Integration by parts gives:

$$ I_1 = \left[ t \sin^{-1}(\sqrt{t}) \right]_0^{\sin^2 x} - \int_0^{\sin^2 x} t \cdot \frac{1}{2\sqrt{t} \sqrt{1 - t}} dt $$

Evaluate the boundary term: at $$t = \sin^2 x$$, $$\sqrt{t} = \sin x$$ (since $$0 \leq x \leq \frac{\pi}{2}$$, $$\sin x \geq 0$$), so $$\sin^{-1}(\sin x) = x$$, and $$t \cdot x = x \sin^2 x$$. At $$t = 0$$, $$\sqrt{t} = 0$$, $$\sin^{-1}(0) = 0$$, and $$t \cdot 0 = 0$$. So the boundary term is $$x \sin^2 x$$.

Simplify the integral:

$$ \int_0^{\sin^2 x} t \cdot \frac{1}{2\sqrt{t} \sqrt{1 - t}} dt = \int_0^{\sin^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt $$

Thus,

$$ I_1 = x \sin^2 x - \int_0^{\sin^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt $$

Next, evaluate $$I_2 = \int_0^{\cos^2 x} \cos^{-1}(\sqrt{t}) dt$$. Again, use integration by parts with $$u = \cos^{-1}(\sqrt{t})$$ and $$dv = dt$$. Then $$v = t$$, and find $$du$$. Set $$w = \sqrt{t}$$, so $$u = \cos^{-1}(w)$$. Then $$\frac{du}{dw} = -\frac{1}{\sqrt{1 - w^2}}$$, and $$\frac{dw}{dt} = \frac{1}{2\sqrt{t}}$$, so $$\frac{du}{dt} = -\frac{1}{\sqrt{1 - w^2}} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{\sqrt{1 - t}} \cdot \frac{1}{2\sqrt{t}}$$. Thus, $$du = -\frac{1}{2\sqrt{t} \sqrt{1 - t}} dt$$.

Integration by parts gives:

$$ I_2 = \left[ t \cos^{-1}(\sqrt{t}) \right]_0^{\cos^2 x} - \int_0^{\cos^2 x} t \cdot \left( -\frac{1}{2\sqrt{t} \sqrt{1 - t}} \right) dt $$

Evaluate the boundary term: at $$t = \cos^2 x$$, $$\sqrt{t} = \cos x$$ (since $$0 \leq x \leq \frac{\pi}{2}$$, $$\cos x \geq 0$$), so $$\cos^{-1}(\cos x) = x$$, and $$t \cdot x = x \cos^2 x$$. At $$t = 0$$, $$\sqrt{t} = 0$$, $$\cos^{-1}(0) = \frac{\pi}{2}$$, but $$t \cdot \frac{\pi}{2} \to 0$$ as $$t \to 0^+$$, so the boundary term is $$x \cos^2 x$$.

Simplify the integral:

$$\int_0^{\cos^2 x} t \cdot \left( -\frac{1}{2\sqrt{t} \sqrt{1 - t}} \right) dt$$ with a negative sign from the formula

The expression becomes:

$$ I_2 = x \cos^2 x + \int_0^{\cos^2 x} \frac{t}{2\sqrt{t} \sqrt{1 - t}} dt = x \cos^2 x + \int_0^{\cos^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt $$

Now, sum $$I_1$$ and $$I_2$$:

$$ I = \left( x \sin^2 x - \int_0^{\sin^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt \right) + \left( x \cos^2 x + \int_0^{\cos^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt \right) $$

$$ I = x \sin^2 x + x \cos^2 x + \left( \int_0^{\cos^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt - \int_0^{\sin^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt \right) $$

Since $$\sin^2 x + \cos^2 x = 1$$, $$x \sin^2 x + x \cos^2 x = x$$. The difference of integrals is:

$$ \int_0^{\cos^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt - \int_0^{\sin^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt = \int_{\sin^2 x}^{\cos^2 x} \frac{\sqrt{t}}{2 \sqrt{1 - t}} dt $$

Thus,

$$ I = x + \frac{1}{2} \int_{\sin^2 x}^{\cos^2 x} \frac{\sqrt{t}}{\sqrt{1 - t}} dt $$

Evaluate the integral $$\int \frac{\sqrt{t}}{\sqrt{1 - t}} dt$$. Use the substitution $$t = \sin^2 \theta$$, so $$dt = 2 \sin \theta \cos \theta d\theta$$, $$\sqrt{t} = \sin \theta$$ (since $$\theta \in [0, \frac{\pi}{2}]$$), and $$\sqrt{1 - t} = \cos \theta$$. Then:

$$ \int \frac{\sqrt{t}}{\sqrt{1 - t}} dt = \int \frac{\sin \theta}{\cos \theta} \cdot 2 \sin \theta \cos \theta d\theta = \int 2 \sin^2 \theta d\theta $$

Since $$2 \sin^2 \theta = 1 - \cos 2\theta$$,

$$ \int 2 \sin^2 \theta d\theta = \int (1 - \cos 2\theta) d\theta = \theta - \frac{1}{2} \sin 2\theta + C = \theta - \sin \theta \cos \theta + C $$

Substitute back: $$\theta = \sin^{-1} \sqrt{t}$$, $$\sin \theta = \sqrt{t}$$, $$\cos \theta = \sqrt{1 - t}$$, so

$$ \int \frac{\sqrt{t}}{\sqrt{1 - t}} dt = \sin^{-1} \sqrt{t} - \sqrt{t} \sqrt{1 - t} + C $$

Now evaluate the definite integral from $$t = \sin^2 x$$ to $$t = \cos^2 x$$:

$$ \left[ \sin^{-1} \sqrt{t} - \sqrt{t} \sqrt{1 - t} \right]_{\sin^2 x}^{\cos^2 x} $$

At upper limit $$t = \cos^2 x$$:

$$ \sin^{-1} (\cos x) - \cos x \cdot \sin x $$

Since $$\sin^{-1} (\cos x) = \sin^{-1} \left( \sin \left( \frac{\pi}{2} - x \right) \right) = \frac{\pi}{2} - x$$ (for $$0 \leq x \leq \frac{\pi}{2}$$), and $$\sqrt{\cos^2 x} = \cos x$$, $$\sqrt{1 - \cos^2 x} = \sin x$$, so:

$$ \frac{\pi}{2} - x - \cos x \sin x $$

At lower limit $$t = \sin^2 x$$:

$$ \sin^{-1} (\sin x) - \sin x \cdot \cos x = x - \sin x \cos x $$

Thus, the definite integral is:

$$ \left( \frac{\pi}{2} - x - \sin x \cos x \right) - \left( x - \sin x \cos x \right) = \frac{\pi}{2} - x - \sin x \cos x - x + \sin x \cos x = \frac{\pi}{2} - 2x $$

Now substitute back into $$I$$:

$$ I = x + \frac{1}{2} \left( \frac{\pi}{2} - 2x \right) = x + \frac{\pi}{4} - x = \frac{\pi}{4} $$

The result is constant and independent of $$x$$. Verifying with specific values:

• At $$x = 0$$, $$I_1 = 0$$, $$I_2 = \int_0^1 \cos^{-1}(\sqrt{t}) dt = \frac{\pi}{4}$$, so $$I = \frac{\pi}{4}$$.
• At $$x = \frac{\pi}{2}$$, $$I_1 = \int_0^1 \sin^{-1}(\sqrt{t}) dt = \frac{\pi}{4}$$, $$I_2 = 0$$, so $$I = \frac{\pi}{4}$$.
• At $$x = \frac{\pi}{4}$$, $$\sin^2 x = \cos^2 x = \frac{1}{2}$$, and $$\sin^{-1}(\sqrt{t}) + \cos^{-1}(\sqrt{t}) = \frac{\pi}{2}$$, so $$I = \int_0^{1/2} \frac{\pi}{2} dt = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$$.

All cases yield $$\frac{\pi}{4}$$. Comparing with the options:

A. $$\frac{\pi}{4}$$

B. 0

C. 1

D. $$-\frac{\pi}{4}$$

Hence, the correct answer is Option A.