Problem Statement
Find the Maclaurin series expansion of $f(x) = \ln(1 + \sin x)$ up to and including the term in $x^3$.
Verified Solution & Marking Scheme
Recall standard series expansions
$$\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)$$
$$\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \dots$$
Substitute u = sin x into logarithmic series
$$u = x - \frac{x^3}{6}$$
$$u^2 = \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + \dots = x^2 + \mathcal{O}(x^4)$$
$$u^3 = x^3 + \mathcal{O}(x^4)$$
Combine terms up to x³
$$\ln(1 + \sin x) = \left( x - \frac{x^3}{6} \right) - \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \dots$$
$$= x - \frac{1}{2}x^2 + \left(-\frac{1}{6} + \frac{1}{3}\right)x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3$$