Using McLaurin’s series expansion, the value of log (sec x) is

McLaurin’s Series Expansion of log(sec x)

To find the McLaurin’s series expansion of a function $f(x)$, we use the formula:

$$f(x)=f(0)+x{f}^{\prime }(0)+\frac{x^2}{2!}{f}^{\prime \prime }(0)+\frac{x^3}{3!}{f}^{\prime \prime \prime }(0)+\frac{x^4}{4!}{f}^{(4)}(0)+\frac{x^5}{5!}{f}^{(5)}(0)+\frac{x^6}{6!}{f}^{(6)}(0)+\dots$$

Let's define our function as $f(x)=\log (\sec x)$ and calculate its derivatives at $x=0$.

**Step 1: Calculate the function value at $x=0$ **

- $f(x)=\log (\sec x)$
- $f(0)=\log (\sec 0)=\log (1)=0$

**Step 2: Calculate the first derivative $f^{\prime }(x)$ and its value at $x=0$ **

- $f^{\prime }(x)=\frac{d}{dx}(\log (\sec x))=\frac{1}{\sec x}\cdot (\sec x\tan x)=\tan x$
- $f^{\prime }(0)=\tan 0=0$

**Step 3: Calculate the second derivative $f^{\prime \prime }(x)$ and its value at $x=0$ **

- $f^{\prime \prime }(x)=\frac{d}{dx}(\tan x)={\sec }^{2}x$
- $f^{\prime \prime }(0)={\sec }^{2}0=1^2=1$

**Step 4: Calculate the third derivative $f^{\prime \prime \prime }(x)$ and its value at $x=0$ **

- $f^{\prime \prime \prime }(x)=\frac{d}{dx}({\sec }^{2}x)=2\sec x(\sec x\tan x)=2{\sec }^{2}x\tan x$
- $f^{\prime \prime \prime }(0)=2{\sec }^{2}0\tan 0=2(1)(0)=0$

**Step 5: Calculate the fourth derivative $f^{(4)}(x)$ and its value at $x=0$ **

- $f^{(4)}(x)=\frac{d}{dx}(2{\sec }^{2}x\tan x)$
- Using the product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$:
- $u=2{\sec }^{2}x⟹u^{\prime }=2(2\sec x(\sec x\tan x))=4{\sec }^{2}x\tan x$
- $v=\tan x⟹v^{\prime }={\sec }^{2}x$
- So, $f^{(4)}(x)=(4{\sec }^{2}x\tan x)(\tan x)+(2{\sec }^{2}x)({\sec }^{2}x)$
- $f^{(4)}(x)=4{\sec }^{2}x{\tan }^{2}x+2{\sec }^{4}x$
- $f^{(4)}(0)=4{\sec }^{2}0{\tan }^{2}0+2{\sec }^{4}0=4(1)(0)+2(1)=2$

**Step 6: Calculate the fifth derivative $f^{(5)}(x)$ and its value at $x=0$ **

- $f^{(5)}(x)=\frac{d}{dx}(4{\sec }^{2}x{\tan }^{2}x+2{\sec }^{4}x)$
- Notice that all terms in $f^{(4)}(x)$ will involve powers of $\tan x$ or $\sec x$ which will result in $\tan x$ in some form after differentiation, or terms that become zero at $x=0$. More formally, differentiating $4{\sec }^{2}x{\tan }^{2}x$ will yield terms with $\tan x$. Differentiating $2{\sec }^{4}x$ will yield $8{\sec }^{4}x\tan x$. Therefore, $f^{(5)}(0)$ will be 0.
- $f^{(5)}(0)=0$

**Step 7: Calculate the sixth derivative $f^{(6)}(x)$ and its value at $x=0$ **

Since $f^{(5)}(0)=0$, we need to find $f^{(6)}(0)$. This can be complex. An alternative approach is to use known series expansions.

We know that $f^{\prime }(x)=\tan x$. The McLaurin series for $\tan x$ is:

$$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots$$

Since $f(x)=\int f^{\prime }(x)dx$, we can integrate the series for $\tan x$ to find the series for $f(x)$:

$$f(x)=\int \left(x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots \right)dx$$

$$f(x)=\frac{x^2}{2}+\frac{x^4}{3\cdot 4}+\frac{2x^6}{15\cdot 6}+\dots +C$$

$$f(x)=\frac{x^2}{2}+\frac{x^4}{12}+\frac{2x^6}{90}+\dots +C$$

$$f(x)=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\dots +C$$

Since $f(0)=0$, the constant of integration $C$ must be 0.

So, the McLaurin's series for $\log (\sec x)$ is:

$$\log (\sec x)=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\dots$$

Step 8: Express the series in factorial form and compare with options

Let's convert the terms to the form $\frac{k{x}^n}{n!}$:

- For the $x^2$ term: $\frac{x^2}{2}=\frac{1\cdot x^2}{2!}$
- For the $x^4$ term: $\frac{x^4}{12}=\frac{2\cdot x^4}{24}=\frac{2x^4}{4!}$
- For the $x^6$ term: $\frac{x^6}{45}=\frac{16\cdot x^6}{720}=\frac{16x^6}{6!}$

Thus, the McLaurin's series expansion for $\log (\sec x)$ is:

$$\log (\sec x)=\frac{x^2}{2!}+\frac{2x^4}{4!}+\frac{16x^6}{6!}+\dots$$

This matches Option 2.

Alternatively, using the derivative method:

Recall the values of derivatives at $x=0$:

- $f(0)=0$
- $f^{\prime }(0)=0$
- $f^{\prime \prime }(0)=1$
- $f^{\prime \prime \prime }(0)=0$
- $f^{(4)}(0)=2$
- $f^{(5)}(0)=0$
- We need $f^{(6)}(0)$. From the series derived by integration: The coefficient of $\frac{x^6}{6!}$ is $f^{(6)}(0)$. We have $\frac{x^6}{45}=\frac{f^{(6)}(0)}{6!}{x}^6$ So, $\frac{1}{45}=\frac{f^{(6)}(0)}{720}$ $f^{(6)}(0)=\frac{720}{45}=16$

Substituting these values into the McLaurin's series formula:

$$f(x)=0+x(0)+\frac{x^2}{2!}(1)+\frac{x^3}{3!}(0)+\frac{x^4}{4!}(2)+\frac{x^5}{5!}(0)+\frac{x^6}{6!}(16)+\dots$$

$$f(x)=\frac{x^2}{2!}+\frac{2x^4}{4!}+\frac{16x^6}{6!}+\dots$$

Both methods yield the same result.