Find $\dfrac{d}{dx}\left( e^{\sin^{-1}x}\sqrt{1-x^{2}}\right).$ (ii) Hence using integration by parts, find $\displaystyle \int x \dfrac{e^{\sin^{-1}x}}{\sqrt{1-x^{2}}}\, dx$.

1. Apply the product rule and chain rule to differentiate the expression.

$\frac{d}{dx}\left( e^{\sin^{-1}x}\sqrt{1-x^{2}}\right) = \left( \frac{d}{dx} e^{\sin^{-1}x} \right) \sqrt{1-x^2} + e^{\sin^{-1}x} \left( \frac{d}{dx} \sqrt{1-x^2} \right)$

$= \left( e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}} \right) \sqrt{1-x^2} + e^{\sin^{-1}x} \left( \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) \right)$

$= e^{\sin^{-1}x} - \frac{x e^{\sin^{-1}x}}{\sqrt{1-x^2}}$

1. Differentiate the expression from part (i) to establish a relationship between the relevant integrals.

$\frac{d}{dx} \left( e^{\sin^{-1}x}\sqrt{1-x^2} \right) = e^{\sin^{-1}x}\left( \frac{1}{\sqrt{1-x^2}} \right)\sqrt{1-x^2} + e^{\sin^{-1}x} \left( \frac{-2x}{2\sqrt{1-x^2}} \right)$
$= e^{\sin^{-1}x} - \frac{x e^{\sin^{-1}x}}{\sqrt{1-x^2}}$
Integrating both sides with respect to $x$ yields:
$e^{\sin^{-1}x}\sqrt{1-x^2} = \int e^{\sin^{-1}x} \, dx - \int x \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} \, dx \quad \text{--- (1)}$

2. Apply integration by parts to the target integral.

Let $I = \displaystyle \int x \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} \, dx$.
Let $u = x \implies \frac{du}{dx} = 1$.
Let $\frac{dv}{dx} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$. Recognizing this is of the standard form $\mathrm{f}'(x)e^{\mathrm{f}(x)}$, we integrate to find $v$:
$v = \int \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} \, dx = e^{\sin^{-1}x}$
Applying the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$:
$I = x e^{\sin^{-1}x} - \int e^{\sin^{-1}x} \, dx \quad \text{--- (2)}$

3. Combine the equations to solve for the target integral.

From equation (1), express $\int e^{\sin^{-1}x} \, dx$ in terms of $I$:
$\int e^{\sin^{-1}x} \, dx = I + e^{\sin^{-1}x}\sqrt{1-x^2}$
Substitute this into equation (2):
$I = x e^{\sin^{-1}x} - \left( I + e^{\sin^{-1}x}\sqrt{1-x^2} \right)$
$2I = x e^{\sin^{-1}x} - e^{\sin^{-1}x}\sqrt{1-x^2}$
$I = \frac{1}{2} e^{\sin^{-1}x} \left( x - \sqrt{1-x^2} \right) + C$