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

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

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

2. Simplify the resulting expression.

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

1. Calculate the derivative from part (i) to establish a relationship between the integrals.

From part (i), we differentiate $e^{\sin^{-1} x} \sqrt{1 - x^2}$ using the product and chain rules:
$\frac{\mathrm{d}}{\mathrm{d}x} \left( e^{\sin^{-1} x} \sqrt{1 - x^2} \right) = e^{\sin^{-1} x} \left( \frac{1}{\sqrt{1 - x^2}} \right) \sqrt{1 - x^2} + e^{\sin^{-1} x} \left( \frac{-2x}{2\sqrt{1 - x^2}} \right)$
$= e^{\sin^{-1} x} - \frac{x e^{\sin^{-1} x}}{\sqrt{1 - x^2}}$
Integrating both sides with respect to $x$, we establish the relationship:
$e^{\sin^{-1} x} \sqrt{1 - x^2} = \int e^{\sin^{-1} x} \, \mathrm{d}x - \int \frac{x e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \, \mathrm{d}x$
Let $I = \int \frac{x e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \, \mathrm{d}x$. Rearranging gives:
$\int e^{\sin^{-1} x} \, \mathrm{d}x = I + e^{\sin^{-1} x} \sqrt{1 - x^2}$

2. Apply integration by parts to the target integral.

Let $I = \int x \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \, \mathrm{d}x$.
We apply integration by parts.
Let $u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1$.
Let $\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \implies v = e^{\sin^{-1} x}$.
Using the integration by parts formula $\int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u$:
$I = x e^{\sin^{-1} x} - \int e^{\sin^{-1} x} \, \mathrm{d}x$

3. Substitute the integral relationship from step 1 into the integration by parts result to solve for I.

Substitute the expression for $\int e^{\sin^{-1} x} \, \mathrm{d}x$ from Step 1 into our equation from Step 2:
$I = x e^{\sin^{-1} x} - \left( I + e^{\sin^{-1} x} \sqrt{1 - x^2} \right)$
$2I = x e^{\sin^{-1} x} - e^{\sin^{-1} x} \sqrt{1 - x^2}$
$I = \frac{1}{2} e^{\sin^{-1} x} \left( x - \sqrt{1 - x^2} \right) + C$
where $C$ is an arbitrary constant of integration.