Integration techniques: EJC 2024 Promo Q6

(a)

$$\begin{aligned} \frac{\,\mathrm{d}}{\,\mathrm{d}\theta}\left(\sin^n\theta\right) &= n\cos\theta\sin^{n-1}\theta. \end{aligned}$$

(b)

$x = 2\cos^2\theta$ $$\frac{\,\mathrm{d} x}{\,\mathrm{d}\theta}=-4\cos\theta\sin\theta$$ $$\begin{aligned} \int x\sqrt{1 - \frac{x}{2}}\,\mathrm{d} x &= \int 2\cos^2\theta\sqrt{1 - \cos^2\theta}\,(-4\cos\theta\sin\theta)\,\mathrm{d}\theta\\ &= \int -8\cos^3\theta\sin\theta\sqrt{\sin^2\theta}\,\mathrm{d}\theta\\ &= \int -8\cos^3\theta\sin^2\theta\,\mathrm{d}\theta\\ &= \int -8\cos\theta\cos^2\theta\sin^2\theta\,\mathrm{d}\theta\\ &= \int -8\cos\theta\left(1 - \sin^2\theta\right)\sin^2\theta\,\mathrm{d}\theta\\ &= 8\int\left(\cos\theta\sin^4\theta - \cos\theta\sin^2\theta\right)\,\mathrm{d}\theta. \quad\text{((Shown))} \end{aligned}$$

(c)

Using part (a), $$\begin{aligned} 8\int\left(\cos\theta\sin^4\theta - \cos\theta\sin^2\theta\right)\,\mathrm{d}\theta &= 8\left(\frac{\sin^5\theta}{5} - \frac{\sin^3\theta}{3}\right) + C. \end{aligned}$$ Since $x = 2\cos^2\theta$, $$\begin{aligned} \therefore\ \sin^2\theta &= 1-\cos^2\theta=1-\frac{x}{2}\\ \therefore\ \sin\theta &= \sqrt{1-\frac{x}{2}} \end{aligned}$$ $$\begin{aligned} \therefore\ \int x\sqrt{1 - \frac{x}{2}}\,\mathrm{d} x &= \frac{8}{5}\left(1 - \frac{x}{2}\right)^{5/2} - \frac{8}{3}\left(1 - \frac{x}{2}\right)^{3/2} + C. \end{aligned}$$