Applications of integration: TJC 2024 Promo Q10

(a)

$$\begin{aligned} \text{Let }\frac{2x}{4+x^2}&=\frac14x\\ 8x&=x(4+x^2)\\ x(x^2-4)&=0\\ x&=0,\ 2\text{ or }-2\ (\text{rejected}) \end{aligned}$$ $$\begin{aligned} \text{Area }R &=\int_0^2\frac{2x}{4+x^2}\,\mathrm{d} x-\int_0^2\frac x4\,\mathrm{d} x\\ &=\left[\ln(4+x^2)\right]_0^2-\left[\frac18x^2\right]_0^2\\ &=\ln8-\ln4-\frac12\\ &=\ln2-\frac12 \end{aligned}$$

(b)

$$x=2\tan\theta,\qquad\frac{\,\mathrm{d} x}{\,\mathrm{d}\theta}=2\sec^2\theta$$ $$\begin{aligned} \int\frac{x^2}{(4+x^2)^2}\,\mathrm{d} x &=\int\frac{4\tan^2\theta}{(4+4\tan^2\theta)^2}(2\sec^2\theta)\,\mathrm{d}\theta\\ &=\int\frac{4\tan^2\theta}{16\sec^4\theta}(2\sec^2\theta)\,\mathrm{d}\theta\\ &=\int\frac{4\tan^2\theta}{8\sec^2\theta}\,\mathrm{d}\theta\\ &=\frac12\int\sin^2\theta\,\mathrm{d}\theta\\ &=\frac14\int(1-\cos2\theta)\,\mathrm{d}\theta\\ &=\frac14\left(\theta-\frac12\sin2\theta\right)+C \end{aligned}$$

Method 1: $$\begin{aligned} \text{The Volume} &=\pi\int_0^2\frac{4x^2}{(4+x^2)^2}\,\mathrm{d} x-\pi\int_0^2\frac{x^2}{16}\,\mathrm{d} x\\ &=\pi\left[\theta-\frac12\sin2\theta\right]_0^{\pi/4} -\pi\left[\frac1{48}x^3\right]_0^2\\ &=\pi\left[\frac\pi4-\frac12\right]-\frac\pi6\\ &=\frac{\pi^2}{4}-\frac{2\pi}{3} \end{aligned}$$

Method 2: $$\begin{aligned} \text{The Volume} &=\pi\int_0^2\frac{4x^2}{(4+x^2)^2}\,\mathrm{d} x -\frac13\pi\left(\frac12\right)^2(2)\\ &=\pi\left[\theta-\frac12\sin2\theta\right]_0^{\pi/4}-\frac\pi6\\ &=\pi\left[\frac\pi4-\frac12\right]-\frac\pi6\\ &=\frac{\pi^2}{4}-\frac{2\pi}{3} \end{aligned}$$