Maclaurin series: VJC 2024 Promo Q9

(a)$\displaystyle\frac1{x^2+a^2}=\frac1{a^2}\left(1+\frac{x^2}{a^2}\right)^{-1}$

$$=\frac1{a^2}\left[1+(-1)\frac{x^2}{a^2}+\frac{(-1)(-2)}{2!}\left(\frac{x^2}{a^2}\right)^2+\ldots\right]$$ $$=\frac1{a^2}\left(1-\frac{x^2}{a^2}+\frac{x^4}{a^4}+\ldots\right)\quad\textbf{OR}\quad\frac1{a^2}-\frac1{a^4}x^2+\frac1{a^6}x^4+\ldots$$ $$\frac{\cos ax}{x^2+a^2}=(\cos ax)\left(1+\frac{x^2}{a^2}\right)^{-1}$$ $$=\frac1{a^2}\left(1-\frac{a^2x^2}{2!}+\frac{a^4x^4}{4!}+\ldots\right)\left(1-\frac{x^2}{a^2}+\frac{x^4}{a^4}+\ldots\right)$$ $$=\frac1{a^2}\left(1-\frac{x^2}{a^2}+\frac{x^4}{a^4}-\frac{a^2x^2}{2!}+\frac{x^4}{2!}+\frac{a^4x^4}{4!}+\ldots\right)$$ $$=\frac1{a^2}\left[1-\left(\frac{a^2}{2}+\frac1{a^2}\right)x^2+\left(\frac1{a^4}+\frac12+\frac{a^4}{24}\right)x^4+\ldots\right]$$ $$=\frac1{a^2}-\frac1{a^2}\left(\frac{a^2}{2}+\frac1{a^2}\right)x^2+\frac1{a^2}\left(\frac1{a^4}+\frac12+\frac{a^4}{24}\right)x^4+\ldots$$ $$=\frac1{a^2}-\left(\frac{a^4+2}{2a^4}\right)x^2+\left(\frac{24+12a^4+a^8}{24a^6}\right)x^4+\ldots$$ $c_1=\dfrac1{a^2}$, $c_2=-\left(\dfrac{a^4+2}{2a^4}\right)$ and $c_3=\dfrac{24+12a^4+a^8}{24a^6}$

9(b)Expansion is valid for $\dfrac{x^2}{a^2}<1$

$x^2<a^2$

$x^2-a^2<0$

$(x-a)(x+a)<0$

$-a<x<a$

9(c)$\displaystyle\int_0^1\frac{\cos2x}{x^2+4}\,dx\approx\int_0^1\left[\frac14-\frac14\left(\frac42+\frac14\right)x^2+\frac14\left(\frac1{16}+\frac12+\frac{16}{24}\right)x^4\right]dx$

$$=\int_0^1\left(\frac14-\frac9{16}x^2+\frac{59}{192}x^4\right)dx=\left[\frac14x-\frac3{16}x^3+\frac{59}{960}x^5\right]_0^1=\frac{119}{960}$$