Figure 2 shows a sketch of the curve with equation $y = ext{sqrt}(3-x)(x + 1)$, $0 \, \leq \,, x \,, \leq 3$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis, and the $y$-axis - Edexcel - A-Level Maths Pure - Question 8 - 2015 - Paper 4

To evaluate the integral, we first compute the differential: $dx = 2 \cos \theta \, d\theta$

Next, substitute $x = 1 + 2 \sin \theta$ into the integrand: $3 - x = 3 - (1 + 2 \sin \theta) = 2 - 2 \sin \theta = 2(1 - \sin \theta)$ $x + 1 = (1 + 2 \sin \theta) + 1 = 2 + 2 \sin \theta = 2(1 + \sin \theta)$ Thus, $\sqrt{(3 - x)(x + 1)} = \sqrt{2(1 - \sin \theta) \cdot 2(1 + \sin \theta)} = 2 \sqrt{(1 - \sin \theta)(1 + \sin \theta)} = 2 \sqrt{1 - \sin^2 \theta} = 2 \cos \theta$

Now, substituting everything into the integral gives: $\int_{0}^{3} \sqrt{(3 - x)(x + 1)} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (2 \cos \theta)(2 \cos \theta) (2 \cos \theta) d\theta = k \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta$ This results in: $\int_{0}^{3}\sqrt{(3-x)(x + 1)} \, dx = 4k \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta$