1. **Problem statement:** (i) Show that $8 \sin^2 x \cos^2 x$ can be written as $1 - \cos 4x$. (ii) Hence find $\int \sin^2 x \cos^2 x \, dx$. 2. **Formula and identities used:** - Double angle identity: $\sin 2x = 2 \sin x \cos x$. - Power-reduction formulas: $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$. - Cosine double angle: $\cos 2\theta = 2 \cos^2 \theta - 1$. 3. **Step (i) Show equivalence:** Start with $8 \sin^2 x \cos^2 x$. Recall $\sin 2x = 2 \sin x \cos x$, so $\sin^2 2x = 4 \sin^2 x \cos^2 x$. Therefore, $8 \sin^2 x \cos^2 x = 2 \sin^2 2x$. Use power-reduction on $\sin^2 2x$: $$\sin^2 2x = \frac{1 - \cos 4x}{2}$$ Multiply by 2: $$2 \sin^2 2x = 2 \times \frac{1 - \cos 4x}{2} = 1 - \cos 4x$$ Hence, $$8 \sin^2 x \cos^2 x = 1 - \cos 4x$$ 4. **Step (ii) Find the integral:** We want to find: $$\int \sin^2 x \cos^2 x \, dx$$ From step (i), $$8 \sin^2 x \cos^2 x = 1 - \cos 4x \implies \sin^2 x \cos^2 x = \frac{1 - \cos 4x}{8}$$ So, $$\int \sin^2 x \cos^2 x \, dx = \int \frac{1 - \cos 4x}{8} \, dx = \frac{1}{8} \int (1 - \cos 4x) \, dx$$ Integrate term-by-term: $$\int 1 \, dx = x$$ $$\int \cos 4x \, dx = \frac{\sin 4x}{4}$$ Therefore, $$\int \sin^2 x \cos^2 x \, dx = \frac{1}{8} \left( x - \frac{\sin 4x}{4} \right) + C = \frac{x}{8} - \frac{\sin 4x}{32} + C$$ **Final answers:** (i) $8 \sin^2 x \cos^2 x = 1 - \cos 4x$ (ii) $\int \sin^2 x \cos^2 x \, dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$