Question

Find the Maclaurin series of the following functions.

$\mathbf{arc}\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{1}\mathbf{+}{\mathbf{u}}^{\mathbf{2}}}$

Short answer

Maclaurin series of $\arctan x\text{is}x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots ..$

Step-by-step solution

The Maclaurin series and the given function.

Function is $\arctan x={\int }_0^x\frac{du}{1+u^2}$

The Maclaurin series of ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{1}\mathbf{+}{\mathbf{u}}^{\mathbf{2}}}$is given as follows:

$$\begin{gathered} {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{1}\mathbf{+}{\mathbf{u}}^{\mathbf{2}}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\left(1-u^2+u^4-u^6+\dots ..\right)\mathit{d}\mathit{u} \\ {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{1}\mathbf{+}{\mathbf{u}}^{\mathbf{2}}}\mathbf{=}{\left(u-\frac{u^3}{3}+\frac{u^5}{5}-\frac{u^7}{7}+\dots ..\right)}_{\mathbf{0}}^{\mathbf{x}} \\ {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{d}\mathbf{u}}{\mathbf{1}\mathbf{+}{\mathbf{u}}^{\mathbf{2}}}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{5}}}{\mathbf{5}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{7}}}{\mathbf{7}}\mathbf{+}\mathbf{\dots } \end{gathered}$$

Now, we have:

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{\dots }\mathbf{.}\mathbf{.}$

Calculation by the Maclaurin series.

Substitute$u^2$ for $x$in expansion of $\frac{1}{1+x}$as follows:

$\begin{array}{rcl}\frac{1}{1+u^2}& =& 1-u^2+{\left(u^2\right)}^2-{\left(u^2\right)}^3+\dots ..\\ \frac{1}{1+u^2}& =& 1-u^2+u^4-u^6+\dots ..\\ & & \end{array}$

Apply integration for $\lim$$0$ to $x$in the above equation as follows:

$$\begin{gathered} {\int }_0^x\frac{du}{1+u^2}={\int }_0^x\left(1-u^2+u^4-u^6+\dots ..\right)du \\ {\int }_0^x\frac{du}{1+u^2}={\left(u-\frac{u^3}{3}+\frac{u^5}{5}-\frac{u^7}{7}+\dots ..\right)}_0^x \\ {\int }_0^x\frac{du}{1+u^2}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots .. \end{gathered}$$

Thus, the Maclaurin series of $\arctan x\text{is}x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots ..$