Question

Find the Maclaurin series $\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\left(x\right)}}$.

Short answer

The answer is$\frac{1}{\sqrt{1+\sin \left(\mathrm{x}\right)}}=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{11x^3}{48}+...$

Step-by-step solution

Given information.

The Maclaurin expansion of the function$\frac{1}{\sqrt{1+\sin \left(\mathrm{x}\right)}}$ is to be evaluated.

Definition of the Maclaurin series

The definition of Maclaurin series is defined as:

$$\begin{gathered} \frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{+}\frac{\mathbf{3}}{\mathbf{8}}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{5}}{\mathbf{16}}{\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.} \\ \mathbf{sin}\left(x\right)\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{5}}}{\mathbf{120}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$

Begin by stating the Maclaurin series.

State the Maclaurin series as:

$$\begin{gathered} \frac{1}{\sqrt{1+\sin \left(\mathrm{x}\right)}}=1-\frac{1}{2}x+\frac{3}{8}{x}^2-\frac{5}{16}{x}^3+... \\ \sin \left(\mathrm{x}\right)=x-\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{120}+... \end{gathered}$$

Combine the two Maclaurin series and evaluate.

Combine the second Maclaurin series in the first and evaluate.

$$\begin{gathered} \frac{1}{\sqrt{1+\sin \left(x\right)}}=1-\frac{1}{2}\left[x-\frac{x^3}{6}+\frac{x^5}{120}+...\right]+\frac{3}{8}{\left[x-\frac{x^3}{6}+...\right]}^2-\frac{5}{16}{\left[x+...\right]}^3+... \\ \frac{1}{\sqrt{1+\sin \left(x\right)}}=1+\frac{-240x+180x^2-110x^3}{480} \\ \frac{1}{\sqrt{1+\sin \left(x\right)}}=1-\frac{x}{2}+\frac{3x^2}{8}+\frac{11x^3}{48}+... \end{gathered}$$

Thus, the final answer is $\frac{1}{\sqrt{1+\sin \left(x\right)}}=1-\frac{x}{2}+\frac{3x^2}{8}+\frac{11x^3}{48}+...$