Finding the Maclaurin series of $e^{\sin x}$ by comparing coefficients I believe I have found a nice way to find the Maclaurin series of $e^{\sin x}$. Please check if there are any mistakes with my working. Is this method well known? 
 A: (Working finds up to $x^5$ and assumes knowledge that the Maclaurin series of $\cos x\approx1-\frac{x^2}{2}+\frac{x^4}{24}$.)
Let $f(x)= e^{\sin x}$, therefore $f'(x)=\cos x*e^{\sin x}=\cos x*f(x)$ by chain rule.
Assume that $f(x)$ has a Maclaurin series and let that series $f(x)=a+bx+cx^2+dx^3+ex^4+fx^5$, therefore $f^{'}(x)=b+2cx+3dx^2+4ex^3+5fx^4$.
Sub in known expressions into $f^{'}(x)=\cos x*f(x)$ to get: $b+2cx+3dx^2+4ex^3+5fx^4=(1-\frac{x^2}{2}+\frac{x^4}{24})(a+bx+cx^2+dx^3+ex^4+fx^5)$.
Expand the right hand side to get: $RHS=a+bx+(c-\frac{1}{2})x^2+(d-\frac{b}{2})x^3+(e-\frac{c}{2}+\frac{a}{24})x^4+(f-\frac{d}{2}+\frac{b}{24})x^5$.
Comparing coefficients of $LHS$ and $RHS$ yields:
$b=a$
$2c=b$
$3d=c-\frac{1}{2}$
$4e=d-\frac{b}{2}$
$5f=e-\frac{c}{2}+\frac{a}{24}$.
When $x=0$, $f(x)=1$ and $f(x)=a$, therefore $a=1$. Which can be used to find:
$b=1$
$c=\frac{1}{2}$
$d=0$
$e=-\frac{1}{8}$
$f=-\frac{1}{15}$.
This means the Maclaurin series of $e^{\sin x}=1+x+\frac{1}{2}x^2-\frac{1}{8}x^4-\frac{1}{15}x^5$ which is indeed to correct series.
