(i)[4]
By differentiating $\dfrac{1}{\cos x}$, demonstrate that the derivative of $\sec x$ is $\sec x \tan x$. Hence demonstrate that if $y = \ln(\sec x + \tan x)$ then $\dfrac{dy}{dx} = \sec x$.
(ii)[4]
Using the substitution $x = (\sqrt{3})\tan \theta$, determine the exact value of $\displaystyle \int_1^3 \dfrac{1}{\sqrt{3 + x^2}}\,dx$, and give your answer as a single logarithm.