Find the Maclaurin series for the following functions.

$\mathbf{ln}\left(\frac{\mathrm{sinx}}{x}\right)$

Function is$\cos [\ln (1+x)].$

The Maclaurin series of $\mathbf{sinx}$is expressed as follows:

$\mathbf{sinx}\mathbf{=}\mathbf{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{5}}}{\mathbf{120}}\mathbf{+}\mathbf{\cdots }\mathbf{\dots }$

The Maclaurin series of$\mathbf{(}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{6}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{180}}\mathbf{-}\mathbf{\dots }\mathbf{\dots }\mathbf{)}$ is expressed as follows:

role="math" localid="1658921544652" $\mathbf{ln}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{\dots }\mathbf{\dots }$

Divide the series sin x and x as follows:

$$\begin{gathered} \left(\frac{\sin x}{x}\right)=\frac{x}{x}-\frac{x^3}{6x}+\frac{x^5}{120x}+........ \\ =1-\frac{x^2}{6}+\frac{x^4}{120}+........ \end{gathered}$$

Take both side in the above equation as follows:

$$\begin{gathered} \ln \left(\frac{\sin x}{x}\right)=In\left(1+\left(-\frac{x^2}{6}+\frac{x^4}{120}\right)+........\right) \\ =In\left(1+\left(-\frac{x^2}{6}+\frac{x^4}{120}\right)+........\right) \end{gathered}$$

Substitute $-\frac{x^2}{6}+\frac{x^4}{120}+\cdots \dots$ for x in expansion of$\ln (1+x)$ as follows:

role="math" localid="1658922803560" $$\begin{gathered} \ln \left(\frac{\sin x}{x}\right)=\left\{\begin{array}{l}-\frac{x^2}{6}+\frac{x^4}{120}+\dots \dots +\frac{{\left(-\frac{x^2}{6}+\frac{x^4}{120}\right)}^2}{2}+\frac{{\left(-\frac{x^2}{6}+\frac{x^4}{120}\right)}^3}{3}\\ \end{array}+........\right. \\ =-\frac{x^2}{6}+\frac{x^4}{180}-........ \end{gathered}$$

Hence, the Maclaurin series of role="math" localid="1658922199244" $\ln \left(\frac{\sin x}{x}\right)$ is role="math" localid="1658922229537" $\left(-\frac{x^2}{6}-\frac{x^4}{180}-\dots \dots \right)$.