Given that \({y = \ln{(\sec{x})}}\), show that \(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 2 \left( \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} \right) \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)\).[3]
Hence or otherwise, obtain the Maclaurin expansion of \({y}\) in terms of \({x}\) up to and including the term in \({x^4}\).[3]
By putting \({x = \frac{\pi}{4}}\), find an approximation for \({\ln{2}}\) in terms of \({\pi}\).[2]
Using your answer to part (b), find an approximation to \(\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{\pi }{10}}{\ln }(\sec x)\mathrm{d}x\). Give your answer correct to 4 significant figures.[1]