Quick Method to Calculate the Maclaurin Series of $\frac{1}{\sqrt{\cos{x}}} $ I am supposed to calculate the maclaurin series for $\frac{1}{\sqrt{\cos{x}}} $ but I can't seem to figure out an efficient way to go about doing this. 
 A: Hint. You may use, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \tag1
$$ and, as $u \to 0$,
$$
\frac 1{\sqrt{1-u}} =1+\frac{u}{2}+\frac{3}{8}u^2+O(u^3) \tag2
$$ giving $$
\begin{align}
\frac 1{\sqrt{\cos x}}&=\frac 1{\sqrt{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}}\\\\
&=1+\frac12\left(\frac{x^2}{2!}-\frac{x^4}{4!}+O(x^6) \right)+\frac{3}{8}\left(\frac{x^2}{2!}-\frac{x^4}{4!}\right)^2+O(x^6)\\\\
&=1+\frac{x^2}{4}+\frac{7 x^4}{96}+O(x^6)
\end{align}
$$ $(1)$ and $(2)$ have been obtained by Taylor expansions near $0$.
A: Just to give another way which will be very close to Olivier Oloa's answer, I would have done $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ Then, long division to get $$\frac{1}{\cos(x)}=1+\frac{x^2}{2}+\frac{5 x^4}{24}+O\left(x^6\right)$$ and then apply $$\sqrt{1+u}=1+\frac{u}{2}-\frac{u^2}{8}+O\left(u^3\right)$$ so $$\frac{1}{\sqrt{\cos{x}}}=1+\frac 12\Big(\frac{x^2}{2}+\frac{5 x^4}{24}+\cdots\Big)-\frac 18\Big(\frac{x^2}{2}+\frac{5 x^4}{24}+\cdots\Big)^2=1+\frac{x^2}{4}+\frac{7 x^4}{96}+\cdots$$
