For $\begin{aligned} x \in & \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ & \log \sec x=y(\text { let })\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{\sec x=e^y \Rightarrow \cos x=e^{-y}}{\cosh y}=\frac{e^y-e^{-y}}{e^y+e^{-y}} \\ & \Rightarrow \quad \frac{\cosh y-\sinh y}{\cosh y+\sinh y}=\frac{e^{-y}}{e^y}=\cos ^2 x \\ & \Rightarrow \quad\left(\frac{\cosh \frac{y}{2}-\sinh \frac{y}{2}}{\cosh \frac{y}{2}+\sinh \frac{y}{2}}\right)^2=\cos ^2 x \\ & \Rightarrow \quad \frac{\cosh \frac{y}{2}-\sinh \frac{y}{2}}{\cosh \frac{y}{2}+\sinh \frac{y}{2}}=\cos x \\ & \Rightarrow \quad \frac{2 \cosh \frac{y}{2}}{2 \sinh \frac{y}{2}}=\frac{1+\cos x}{1-\cos x}=\cot ^2 \frac{x}{2}\end{aligned}$
$\begin{array}{rlrl}\Rightarrow & & \operatorname{coth} \frac{y}{1} & =\cot ^2 \frac{x}{2}=\operatorname{cosec}^2 \frac{x}{2}-1 \\ \Rightarrow & \frac{y}{2} & =\operatorname{coth}^{-1}\left(\operatorname{cosec}^2 \frac{x}{2}-1\right) \\ \Rightarrow & y & =2 \operatorname{coth}^{-1}\left(\operatorname{cosec}^2 \frac{x}{2}-1\right)\end{array}$