Let,$ y = log[ 1 \; + \; sin \; x] \\ \; \\ \therefore \frac{dy}{dx} \; = \; \frac{1}{1+sin\;x} \frac{d}{dx} (1+sin\;x) \; = \; \frac{cos\;x}{1+sin\;x} \\ \; \\ \frac{dy}{dx} \; = \; \Big[\frac{1}{1+sin\;x} \Big] cos\;x \\ \; \\ \; \\ \left[ \begin{array}{ccc} \\ \; \dfrac{1}{ 1 + \bigg( x-\frac{x^3}{3!}+\ldots \bigg)} \\ \; \end{array} \right] \bigg[ 1 - \frac{x^2}{2!} + \ldots \bigg] \; \; \left\{ \begin{array}{ccc} \because sinx \; = \; x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \& \\ cosx \; = \; 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \\ Neglecting \; Higher \; Order \; Terms \end{array} \right\} \\ \; \\ = \Bigg[ 1 - \bigg( x-\frac{x^3}{3!}+\ldots \bigg) + \bigg( x-\frac{x^3}{3!}+\ldots \bigg)^2 \Bigg] \bigg[ 1 - \frac{x^2}{2!} + \ldots \bigg] \\ \Big\{\because \frac{1}{1+X} \; = \; 1-X+X^2-X^3+\ldots \Big\} \\ \; \\ \; \\ = \Bigg[ 1 - \bigg( x-\frac{x^3}{3!}+\ldots \bigg) + x^2 - \frac{2x^4}{3!} + \frac{x^6}{(3!)^2} + \ldots \Bigg] \bigg[ 1 - \frac{x^2}{2!} + \ldots\bigg] \\ \; \\ = \Bigg[ 1 - x+\frac{x^3}{3!} + x^2 - \frac{2x^4}{3!} + \frac{x^6}{(3!)^2} + \ldots \Bigg] \bigg[ 1 - \frac{x^2}{2!} + \ldots \bigg] \\ \; \\ = [1-x+x^2+\ldots] \Big[ 1 - \frac{x^2}{2!} + \ldots \Big] \; \; \{ \because 2!=2 \; and \; neglecting \; higher \; order \; terms \; \} \\ \; \\ = 1-x+x^2 \; - \; \frac{x^2}{2}( 1-x+x^2+\ldots ) \; + \; \ldots \\ \; \\ = 1-x+x^2 - \frac{x^2}{2} + \frac{x^3}{2} \; \ldots \\ \; \\ = 1-x+ \frac{x^2}{2} + \ldots \\ \; \\ On \; integrating, \; we \; get \\ \; \\ \therefore y=x-\frac{x^2}{2}+ \frac{x^3}{6}+\ldots \\ \; \\ \; \\ Hence \; log[1+sinx] \; = \; x-\frac{x^2}{2} +\frac{x^3}{6}+\ldots $