Maclaurin series: DHS 2024 Promo Q8

(i) $$\begin{aligned} \cos3x&=1-\frac{(3x)^2}{2!}+\frac{(3x)^4}{4!}+\cdots =1-\frac92x^2+\frac{27}{8}x^4+\cdots\\ \ln(1+\cos3x) &=\ln\left(1+\left(1-\frac92x^2+\frac{27}{8}x^4+\cdots\right)\right)\\ &=\ln2\left(1-\frac94x^2+\frac{27}{16}x^4+\cdots\right)\\ &=\ln2+\ln\left(1-\frac94x^2+\frac{27}{16}x^4+\cdots\right)\\ &=\ln2+\left(-\frac94x^2+\frac{27}{16}x^4+\cdots\right) -\frac{\left(-\frac94x^2+\frac{27}{16}x^4+\cdots\right)^2}{2}+\cdots\\ &=\ln2-\frac94x^2+\frac{27}{16}x^4-\frac{81}{32}x^4+\cdots\\ &=\ln2-\frac94x^2-\frac{27}{32}x^4+\cdots \end{aligned}$$ (ii) $$\begin{aligned} \int_0^{0.5}x\ln(1+\cos3x)\,dx &=\int_0^{0.5}x\left(\ln2-\frac94x^2-\frac{27}{32}x^4+\cdots\right)dx\\ &=\int_0^{0.5}\left((\ln2)x-\frac94x^3-\frac{27}{32}x^5+\cdots\right)dx\\ &=\left[(\ln2)\frac{x^2}{2}-\frac9{16}x^4-\frac9{64}x^6+\cdots\right]_0^{0.5}\\ &\approx0.04929\quad(5\text{ d.p.}) \end{aligned}$$ (iii) Using GC, $\displaystyle\int_0^{0.5}x\ln(1+\cos3x)\,dx=0.04900\ (5\text{ d.p.})$

(iv)

From the diagram, it can be seen that the graphs of $y=x\ln(1+\cos3x)$ and $y=x\left(\ln2-\dfrac94x^2-\dfrac{27}{32}x^4\right)$ are close to each other mostly from $x=0$ to $x=0.5$. Hence, the approximated value of $\displaystyle\int_0^{0.5}x\left(\ln2-\frac94x^2-\frac{27}{32}x^4+\cdots\right)dx$ from (ii) is approximately equal to the actual value of $\displaystyle\int_0^{0.5}x\ln(1+\cos3x)dx=0.04900\ (5\text{ d.p.})$ in (iii).