Question

$sin\left[ln\right(1+x\left)\right]$

Short answer

$\sin \left[\ln \right(1+x\left)\right]=x-\frac{x^2}{2}+\frac{x^3}{6}+\dots$

Step-by-step solution

Given information

The Maclaurin expansion of the function$\sin \left[\ln \right(1+x\left)\right]$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined,

$sin\left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$\sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+\dots$

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

Part-B Step 3: Use the first Maclaurin series to evaluate

Substitute $x$with $\left[x-\frac{x^2}{2}+\frac{x^3}{3}+\dots \right]$in the first series and evaluate.

$\sin \left[\ln \right(1+x\left)\right]=\left[x-\frac{x^2}{2}+\frac{x^3}{3}+\dots \right]-\frac{{\left[x-\frac{x^2}{2}+\dots \right]}^3}{6}+\dots$

$\sin \left[\ln \right(1+x\left)\right]=\left[x-\frac{x^2}{2}+\frac{x^3}{3}+\dots \right]-\frac{x^3-\frac{3x^4}{2}+\frac{3x^5}{4}-\frac{x^6}{8}+\dots }{6}+\dots$

$\sin \left[\ln \right(1+x\left)\right]=x-\frac{x^2}{2}+\frac{x^3}{6}+\dots$

Thus, the final answer is$\sin \left[\ln \right(1+x\left)\right]=x-\frac{x^2}{2}+\frac{x^3}{6}+\dots$