Question

${\mathit{e}}^{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}$

Short answer

$e^{\sin \left(x\right)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+\dots$

Step-by-step solution

Given information.

The Maclaurin expansion of the function $e^{\sin x}$ is to be evaluated.

Definition of Maclaurin series.

The definition of the Maclaurin series is defined.

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

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

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

State the Maclaurin series.

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

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

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

Substitute $x$with $\left[x-\frac{x^3}{6}+\frac{x^5}{120}+\dots \right]$in the second series and evaluate.

$e^{\sin \left(x\right)}=1+\left[x-\frac{x^3}{6}+\frac{x^5}{120}+\dots \right]+\frac{{\left[x-\frac{x^3}{6}+\frac{x^3}{120}+\dots \right]}^2}{2}+\frac{{\left[x-\frac{x^3}{6}+\frac{x^0}{120}+\dots \right]}^3}{6}+\frac{{\left[x-\frac{x^3}{6}+\frac{x^3}{120}+\dots \right]}^4}{24}+\dots$

$e^{\sin \left(x\right)}=1+\frac{155520x}{155520}+\frac{77760x^2}{155520}-\frac{19440x^4}{155520}+\dots$

$e^{\sin \left(x\right)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+\dots$

Thus, the final answer is$e^{\sin \left(x\right)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+\dots$