Section A: Pure Mathematics [40 marks]

1. Apply the chain rule to differentiate the function

Let $y = \tan(x^3)$.
Using the chain rule, we differentiate the outer trigonometric function and multiply it by the derivative of the inner polynomial:
$\frac{dy}{dx} = \sec^2(x^3) \cdot \frac{d}{dx}(x^3)$
$\frac{dy}{dx} = 3x^2 \sec^2(x^3)$

1. State the derivative from part (i) to identify parts for integration.

From part (i), we have $\frac{d}{dx}(\tan(x^3)) = 3x^2 \sec^2(x^3)$.
We split the integrand to prepare for integration by parts:
$\int 6x^5 \sec^2(x^3) \, dx = \int (2x^3) \cdot (3x^2 \sec^2(x^3)) \, dx$

2. Apply integration by parts using the identified components.

Let $u = 2x^3 \implies \frac{du}{dx} = 6x^2$
Let $\frac{dv}{dx} = 3x^2 \sec^2(x^3) \implies v = \tan(x^3)$
Using the integration by parts formula $\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$:
$\int (2x^3) \cdot (3x^2 \sec^2(x^3)) \, dx = 2x^3 \tan(x^3) - \int \tan(x^3) \cdot 6x^2 \, dx$

3. Integrate the remaining term using the standard result for $\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)} \, dx$.

$= 2x^3 \tan(x^3) - 2 \int \frac{3x^2 \sin(x^3)}{\cos(x^3)} \, dx$
Recognising that the numerator is closely related to the derivative of the denominator (since $\frac{d}{dx}(\cos(x^3)) = -3x^2 \sin(x^3)$):
$= 2x^3 \tan(x^3) + 2 \int \frac{-3x^2 \sin(x^3)}{\cos(x^3)} \, dx$
$= 2x^3 \tan(x^3) + 2 \ln|\cos(x^3)| + C$