[Cambridge/9233/2005]

1. Differentiate the given function using the chain rule.

Let $y = \sin(x^2)$.
Using the chain rule,
$\frac{dy}{dx} = \cos(x^2) \cdot \frac{d}{dx}(x^2)$
$\frac{dy}{dx} = 2x \cos(x^2)$

1. Rewrite the integral to incorporate the derivative from part (i).

From part (i), we have $\frac{d}{dx}(\sin(x^2)) = 2x \cos(x^2)$.

To make use of this, we can split $x^3$ into $x^2 \cdot x$ and rewrite the integral as:
$\int x^3 \cos(x^2) \, dx = \frac{1}{2} \int x^2 \big(2x \cos(x^2)\big) \, dx$

2. Apply integration by parts.

Let $u = x^2$ and $\frac{dv}{dx} = 2x \cos(x^2)$.
Then $\frac{du}{dx} = 2x$ and $v = \sin(x^2)$.

Using the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$:
$\frac{1}{2} \int x^2 \big(2x \cos(x^2)\big) \, dx = \frac{1}{2} \left[ x^2 \sin(x^2) - \int 2x \sin(x^2) \, dx \right]$

3. Integrate the remaining term to find the final answer.

Notice that $\frac{d}{dx}(\cos(x^2)) = -2x \sin(x^2)$, so $\int 2x \sin(x^2) \, dx = -\cos(x^2)$.

Substituting this back:
$\frac{1}{2} \left[ x^2 \sin(x^2) - (-\cos(x^2)) \right] + C$
$= \frac{1}{2}x^2 \sin(x^2) + \frac{1}{2}\cos(x^2) + C$