Solve: d/dx {∫_0^x sin²(t) (sin⁻¹(√t)) dt}

Steps to Solve

1. Recognize the Integral Expression The function we want to differentiate is given by the expression $$ F(x) = \int_0^x \sin^2(t) \cdot \sin^{-1}(\sqrt{t}) , dt. $$

2. Apply the Fundamental Theorem of Calculus According to the Fundamental Theorem of Calculus, if we have $$ F(x) = \int_a^x f(t) , dt, $$ then $$ \frac{d}{dx} F(x) = f(x). $$

Here, we need to differentiate $F(x)$: $$ \frac{d}{dx} F(x) = \sin^2(x) \cdot \sin^{-1}(\sqrt{x}). $$

3. Differentiate using the Product Rule Since $F(x)$ is the product of two functions, we must use the product rule: If $u = \sin^2(x)$ and $v = \sin^{-1}(\sqrt{x})$, then $$ \frac{d}{dx}(uv) = u'v + uv'. $$

Compute $u'$ and $v'$:

• For $u = \sin^2(x)$, using the chain rule, $$ u' = 2\sin(x)\cos(x) = \sin(2x). $$

• For $v = \sin^{-1}(\sqrt{x})$, we apply the chain rule: $$ v' = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{1 - x^2}}. $$

Now, apply the product rule: $$ \frac{d}{dx} F(x) = \sin^2(x) \cdot \frac{1}{2\sqrt{1 - x}} + \sin(2x) \cdot \sin^{-1}(\sqrt{x}). $$

4. Simplification At this stage, we evaluate whether any of the options provided in the question match this derivative.