Integration techniques: ASRJC 2024 Promo Q9

(a)

$$\begin{aligned} \int\frac{x+2}{x^2+2x-3}\,\mathrm{d} x &=\int\frac12\left(\frac{2x+2}{x^2+2x-3}\right) +\frac1{(x+1)^2-(2)^2}\,\mathrm{d} x\\ &=\frac12\ln|x^2+2x-3|+\frac1{2(2)}\ln\left|\frac{(x+1)-2}{(x+1)+2}\right|+C\\ &=\frac12\ln|x^2+2x-3|+\frac14\ln\left|\frac{x-1}{x+3}\right|+C \end{aligned}$$

Alternative Method $$\begin{aligned} \int\frac{x+2}{x^2+2x-3}\,\mathrm{d} x &=\frac14\int\left(\frac3{x-1}+\frac1{x+3}\right)\,\mathrm{d} x\\ &=\frac34\ln|x-1|+\frac14\ln|x+3|+C \end{aligned}$$

(b)

$$\begin{aligned} \int2x\sin x\,\mathrm{d} x&=2x(-\cos x)-\int2(-\cos x)\,\mathrm{d} x\\ &=-2x\cos x+\int2\cos x\,\mathrm{d} x\\ &=-2x\cos x+2\sin x+C \end{aligned}$$

(c)

For $x=3\sin\theta$ $$\frac{\,\mathrm{d} x}{\,\mathrm{d}\theta}=3\cos\theta$$ $$\begin{aligned} \int_0^{3/2}\sqrt{9-x^2}\,\mathrm{d} x &=\int_0^{\pi/6}\sqrt{9-(3\sin\theta)^2}\left(\frac{\,\mathrm{d} x}{\,\mathrm{d}\theta}\right)\,\mathrm{d}\theta\\ &=\int_0^{\pi/6}3\sqrt{1-\sin^2\theta}(3\cos\theta)\,\mathrm{d}\theta\\ &=\int_0^{\pi/6}9\cos^2\theta\,\mathrm{d}\theta\\ &=9\int_0^{\pi/6}\frac{\cos2\theta+1}{2}\,\mathrm{d}\theta\\ &=\frac92\left[\frac{\sin2\theta}{2}+\theta\right]_0^{\pi/6}\\ &=\frac94\sin\frac\pi3+\frac92\left(\frac\pi6\right)-0-0\\ &=\frac94\left(\frac{\sqrt3}{2}\right)+\frac34\pi\\ &=\frac98\sqrt3+\frac34\pi \end{aligned}$$