We have

$\log (1+\sin x) = \sin x - \dfrac {(\sin x)^2} {2}+ \dfrac {(\sin x)^3}{3} - \cdots $

$= \left ( x- \dfrac {x^3}{6}+ \cdots \right ) - \dfrac {1}{2} \left ( \left ( x- \dfrac {x^3}{6}+ \cdots \right ) ^2\right ) + \dfrac {1}{3} \left ( x- \dfrac {x^3}{6} + \cdots \right )^3 + \cdots $

$ = x- \dfrac {x^3}{6}+ \cdots - \dfrac {1}{2} \left ( x^2 - \dfrac {2x^4}{6}+ \cdots \right )+ \dfrac {1}{3} (x^3 + \cdots ) - \dfrac {1}{4} (x^4 + \cdots)$

$\therefore \log (1+\sin x) = \left [ x- \dfrac {x^2}{2}+\dfrac {x^3}{6} + \cdots \right ]$