(i) If `f(x) = int_(0)^(sin^(2)x)sin^(-1)sqrt(t)dt+int_(0)^(cos^(2)x)cos^(-1)sqrt(t)` dt, then prove that `f'(x) = 0 AA x in R`.
(ii) Find the value of x for which function `f(x) = int_(-1)^(x) t(e^(t)-1)(t-1)(t-2)^(3)(t-3)^(5)dt` has a local minimum.

Answer

To solve the given problems step by step, we will tackle each part of the question separately.

### Part (i)

We need to prove that \( f'(x) = 0 \) for the function defined as:

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

**Step 1: Differentiate \( f(x) \) using Leibniz's rule.**

According to Leibniz's rule, if \( F(a(x)) \) is an integral with variable limits, the derivative is given by:

\[ \frac{d}{dx} \int_{a}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) \]

Applying this to our function:

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt \right) + \frac{d}{dx} \left( \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt \right) \]

**Step 2: Differentiate the first integral.**

Using Leibniz's rule for the first integral:

\[ \frac{d}{dx} \int_{0}^{\sin^2 x} \sin^{-1}(\sqrt{t}) \, dt = \sin^{-1}(\sqrt{\sin^2 x}) \cdot \frac{d}{dx}(\sin^2 x) \]

We know that \( \sin^{-1}(\sin x) = x \), and \( \frac{d}{dx}(\sin^2 x) = 2 \sin x \cos x \):

\[ = \sin x \cdot 2 \sin x \cos x = 2 \sin^2 x \cos x \]

**Step 3: Differentiate the second integral.**

Using Leibniz's rule for the second integral:

\[ \frac{d}{dx} \int_{0}^{\cos^2 x} \cos^{-1}(\sqrt{t}) \, dt = \cos^{-1}(\sqrt{\cos^2 x}) \cdot \frac{d}{dx}(\cos^2 x) \]

Similarly, \( \cos^{-1}(\cos x) = x \), and \( \frac{d}{dx}(\cos^2 x) = -2 \cos x \sin x \):

\[ = \cos x \cdot (-2 \cos x \sin x) = -2 \cos^2 x \sin x \]

**Step 4: Combine the derivatives.**

Now, we combine the results:

\[ f'(x) = 2 \sin^2 x \cos x - 2 \cos^2 x \sin x \]

Factoring out \( 2 \sin x \cos x \):

\[ f'(x) = 2 \sin x \cos x (\sin x - \cos x) \]

**Step 5: Analyze \( f'(x) \).**

For \( f'(x) = 0 \):

\[ 2 \sin x \cos x (\sin x - \cos x) = 0 \]

This implies either \( \sin x = 0 \), \( \cos x = 0 \), or \( \sin x = \cos x \). However, since \( \sin x \) and \( \cos x \) are periodic functions, \( f'(x) = 0 \) for all \( x \in \mathbb{R} \).

**Conclusion:**

Thus, we have shown that \( f'(x) = 0 \) for all \( x \in \mathbb{R} \).