Match two series without matching their cubic terms

Set $y=\ln(1+2\sin x)$, so $e^y=1+2\sin x$. Repeated differentiation gives

$e^y y^{\prime}=2\cos x,\quad e^y[y^{\prime\prime}+(y^{\prime})^2]=-2\sin x,$

$\begin{aligned}&e^y[y^{\prime\prime\prime}+3y^{\prime}y^{\prime\prime}+(y^{\prime})^3]\\ & =-2\cos x.\end{aligned}$

At zero, $y=0$, $y^{\prime}=2$, $y^{\prime\prime}=-4$; then $y^{\prime\prime\prime}-24+8=-2$, so $y^{\prime\prime\prime}=14$.

Direct differentiation gives

$f^{\prime}(x)=\frac{2\cos x}{1+2\sin x},\quad f^{\prime\prime}(x)=-\frac{4+2\sin x}{(1+2\sin x)^2},\quad f^{\prime\prime\prime}(x)=\frac{(14+4\sin x)\cos x}{(1+2\sin x)^3}.$

These yield $f(0)=0$, $f^{\prime}(0)=2$, $f^{\prime\prime}(0)=-4$, $f^{\prime\prime\prime}(0)=14$, agreeing with Method 1. Therefore $f(x)=2x-2x^2+7x^3/3+O(x^4)$.

The standard product is

$\begin{aligned}&e^{ax}\sin(nx)\\ & =nx+anx^2\\ &{}+\left(\frac{a^2n}{2}-\frac{n^3}{6}\right)x^3\\ &{}+O(x^4).\end{aligned}$

Matching $2x-2x^2$ gives $n=2$ and $a=-1$. The next coefficient is $1-8/6=-1/3$, so its third nonzero term is $-x^3/3$. It need not equal the logarithm’s third term $7x^3/3$.