$If \; y= log \; sec \; x , then, \\ {dy \over dx} = tan x = x+ \frac {x^3} 3 + \frac {2x^5}{15}+ \dots \\ \therefore integrating, \\ y = \; a_0 + \frac{x^2}2 + \frac13. \frac{x^4}4 + \frac2{15} .\frac {x^6}6 + \dots \\ When \; x=0, \; y=0, \; a_0 \; becomes \; 0 \\ \therefore y= \frac{x^2}2 + \frac{x^4}{12} +\frac{x^6}{45} + \dots \\ Hence \; proved.$