Let \(y = log(1 + sinx)\)
\(y(0) = 0\)
We have,
\(y_1 = \frac{\cos x}{1 + \sin x}\)
\(y_1(0) = 1\)
Now,
\(y_1 = \cfrac{\sin (\frac \pi 2 - x)}{1 + \cos (\frac \pi 2-x)}\)
\(= \cfrac{2\sin \frac 12 (\frac \pi 2 - x)\cos \frac 12 (\frac \pi 2 - x)}{2\cos^2 \frac 12 (\frac \pi 2 - x)}\)
\(\therefore y_1 = \tan (\frac \pi 4 - \frac x2)\)
\(\therefore y_2 = -\frac 12 \sec^2 (\frac \pi4 - \frac x2)\)
\(= -\frac 12\left[1 + \tan^2(\frac \pi 2 - \frac x2)\right]\)
\(y_2 = \frac{-1}2 (1 + y_1^2)\)
\(y_2(0) = \frac{-1}2(1 + 1) = -1\)
\(y_3 = -\frac 12 2y_1y_2 = - y_1y_2\)
This gives, \(y_3(0) = -1 (-1) = 1\)
\(y_4 = [y_1y_3 +y_2^2]\)
\(y_4(0) = - [1 \times 1 + (-1)^2] = -2\)
\(y_5 = -[y_1y_4 + y_2 y_3 + 2y_2y_3]\)
\(= - (y_1 y_4 + 3y_2 y_3)\)
which gives,
\(y_5(0 ) = - [1 (-2) + 3-1) ] = 5\)
Therefore by Maclaurin's theorem, we have
\(y = y(0) + xy_1 (0) + \frac{x^2}{2!}y_2(0) + ...\)
\(\log(1 + \sin x)= 0 + x.1+ \frac{x^2}{2!}(-1) + \frac{x^3}{3!} + \frac{x^3}{4!}(-2)+ \frac{x^5}{5!}(5) + .....\)
\(= x - \frac{x^2}{2!} + \frac{x^3}{3!} - \frac{2x^4}{4!}+ \frac{5x^5}{5!}+.....\)
