Let, $ y\;=\;log(sec \; x) \; \; \therefore \dfrac{dy}{dx} \; = \; \dfrac{1}{sec \;x} sec\;xtan\;x \; = \; tan\;x \\ \; \\ By \; using \; standard \; series \; of \; tan x, \\ \; \\ \therefore \dfrac{dy}{dx} \; = \; tan\;x \; = \; x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\ldots \\ \; \\ On \; integrating, \\ \; \\ \; \\ \therefore y\;=\;log(sec \; x) \; = \; \dfrac{x^2}{2}+\dfrac{x^4}{12}+\dfrac{2x^6}{6*15}+\ldots \; = \; \dfrac{x^2}{2} + \dfrac{x^4}{12} + \dfrac{x^6}{45}+\ldots Hence \; Proved. $