Recover a logarithmic series from a differential identity

$y^\prime=\cos x/(1+\sin x)$ and $e^y=1+\sin x$, so $e^yy^\prime=\cos x$. Differentiating gives $e^yy^{\prime\prime}+e^y(y^\prime)^2=-\sin x=1-e^y$. Rearranging proves $e^yy^{\prime\prime}=1-e^y[(y^\prime)^2+1]$.

$y(0)=0$ and $y^\prime(0)=1$. The identity gives $y^{\prime\prime}(0)=1-(1+1)=-1$. Thus $y=x-x^2/2+O(x^3)$.

$\sin x=x+O(x^3)$ and $\ln(1+u)=u-u^2/2+O(u^3)$. Substituting $u=\sin x$ gives $y=\sin x-(\sin x)^2/2+O(x^3)=x-x^2/2+O(x^3)$.