8. Recall that $\cos x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}$.
(a) Write a power series for $g(x)=x^2\cos \left(x^2\right)$.
(b) Write a power series for $g^{\prime }(x)$. (You should not need the product rule or chain rule.)
(c) Write a power series for $G(x)=\int x^2\cos \left(x^2\right)dx$, given $F(0)=3$. (You should not need any special methods of integration.)

Step 1 ：Given the power series for $\cos x$ as $\cos x=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^{2n}}{(2n)!}$, we can substitute $x^2$ into the power series for $\cos x$ and then multiply the result by $x^2$ to get the power series for $g(x)=x^2\cos \left(x^2\right)$.

Step 2 ：The power series for $g(x)=x^2\cos \left(x^2\right)$ is $$g(x)=x^2(-\frac{x^{36}}{6402373705728000}+\frac{x^{32}}{20922789888000}-\frac{x^{28}}{87178291200}+\frac{x^{24}}{479001600}-\frac{x^{20}}{3628800}+\frac{x^{16}}{40320}-\frac{x^{12}}{720}+\frac{x^8}{24}-\frac{x^4}{2}+1)$$

Step 3 ：We can differentiate the power series for $g(x)$ term by term to get the power series for $g^{\prime }(x)$.

Step 4 ：The power series for $g^{\prime }(x)$ is $$g^{\prime }(x)=x^2(-\frac{x^{35}}{177843714048000}+\frac{x^{31}}{653837184000}-\frac{x^{27}}{3113510400}+\frac{x^{23}}{19958400}-\frac{x^{19}}{181440}+\frac{x^{15}}{2520}-\frac{x^{11}}{60}+\frac{x^7}{3}-2x^3)+2x(-\frac{x^{36}}{6402373705728000}+\frac{x^{32}}{20922789888000}-\frac{x^{28}}{87178291200}+\frac{x^{24}}{479001600}-\frac{x^{20}}{3628800}+\frac{x^{16}}{40320}-\frac{x^{12}}{720}+\frac{x^8}{24}-\frac{x^4}{2}+1)$$

Step 5 ：We can integrate the power series for $g(x)$ term by term and then add the constant $F(0)=3$ to get the power series for $G(x)=\int x^2\cos \left(x^2\right)dx$.

Step 6 ：The power series for $G(x)=\int x^2\cos \left(x^2\right)dx$ with $F(0)=3$ is $$G(x)=-\frac{x^{39}}{249692574523392000}+\frac{x^{35}}{732297646080000}-\frac{x^{31}}{2702527027200}+\frac{x^{27}}{12933043200}-\frac{x^{23}}{83462400}+\frac{x^{19}}{766080}-\frac{x^{15}}{10800}+\frac{x^{11}}{264}-\frac{x^7}{14}+\frac{x^3}{3}+3$$