Maclaurin expansion of log(1+sin x)

Let f(x) = log(1 + sin x). Use Maclaurin's series f(x) = f(0) + x f′(0) + (x²/2!)f″(0) + ...

f′  = cos x/(1 + sin x)
f″  = [−sin x(1+sin x) − cos²x]/(1+sin x)² = −1/(1 + sin x)
f‴  = cos x/(1 + sin x)²
f⁗  = [−sin x(1+sin x) − 2cos²x]/(1+sin x)³

At x = 0:

f(0) = 0,  f′(0) = 1,  f″(0) = −1,  f‴(0) = 1,  f⁗(0) = −2

Therefore

f(x) = 0 + x − x²/2! + x³/3! − 2x⁴/4! + ...

Answer: log(1 + sin x) = x − x²/2 + x³/6 − x⁴/12 + ...

(Check: substituting u = sin x = x − x³/6 in log(1+u) = u − u²/2 + u³/3 − u⁴/4 gives the same series.)