Question

$\mathbf{In}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}$Hints:Method1:Write$\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{u}$;use the series you know for $\mathbf{In}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{u}\mathbf{)}$;replace u by the Maclaurin series for$\mathbf{}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}$

Method2:$\mathbf{}\mathbf{In}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{}$$\mathbf{-}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\mathbf{tan}\mathbf{}\mathbf{udu}$Use the series of Example 2 in method B.

Short answer

$\mathrm{In}\left(\cos \right(\mathrm{x}\left)\right)=$$\frac{-\left(x^2\right)}{2}$localid="1658337115294" $\frac{-\left(x^4\right)}{12}$$\frac{-\left(x^6\right)}{45}$$\frac{}{}$....

Step-by-step solution

Given information

The Maclaurin expansion of the function$In\cos x$is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$\mathit{I}\mathit{n}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}{\frac{\mathbf{x}}{\mathbf{3}}}^{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

$\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

$\mathit{I}\mathit{n}\mathbf{}\mathbf{\left(}\mathit{c}\mathit{o}\mathit{s}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathit{I}\mathit{n}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{6}}}{\mathbf{720}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}$

Begin by stating the Maclaurin series

State the Maclaurin series.

$In\left(1+x\right)=$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$$+$......

$\cos \left(x\right)=1-$$\frac{x^2}{2}+\frac{x^4}{24}$$\frac{-x^6}{720}+\dots$

$In\left(\cos \left(x\right)\right)=$$In(1-\frac{x^2}{2}+\frac{x^4}{24}$$-\frac{x^6}{720}+\dots$$)$

Rewrite the expression in a suitable form.

Rewrite the expression $(1-\frac{x^2}{2}+\frac{x^4}{24}+.....)$in such a way that it resembles In (1+x)

Make appropriate substitution in the Maclaurin series and evaluate

Replace x with the expression$\left[-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+....\right]$in the first defined Maclaurin series.

$$\begin{gathered} In\left(\cos \right(x\left)\right)=In(1+\left[-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+....\right]) \\ =\left[-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+....\right]-\frac{{\left[-{\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{x^4}{24}}+...\right]}^2}{2}+{\frac{\left[-{\displaystyle \frac{x^2}{2}}+.....\right]}{3}}^3+.... \end{gathered}$$

$$\begin{gathered} In\left(\cos \right(x\left)\right)=-\frac{\left(2880x^2+480x^4+128x^6\right)}{5760}+..... \\ In(\cos (x\left)\right)=-\frac{\left(x^2\right)}{2}-\frac{\left(x^4\right)}{12}-\frac{\left(x^6\right)}{45}-....... \end{gathered}$$

Thus,the final answer is$In(\cos (x\left)\right)=-\frac{\left(x^2\right)}{2}-\frac{\left(x^4\right)}{12}-\frac{\left(x^6\right)}{45}-.....$