Answer
We show the following:
$T \approx 2\pi \sqrt {\dfrac{L}{g}} \left( {1 + \dfrac{{{\theta ^2}}}{{16}}} \right)$
Work Step by Step
Recall from Example 12, we have
$\dfrac{1}{{\sqrt {1 - {k^2}{{\sin }^2}t} }} = 1 + \dfrac{1}{2}{k^2}{\sin ^2}t + \dfrac{{1\cdot3}}{{2\cdot4}}{k^4}{\sin ^4}t + \dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}{k^6}{\sin ^6}t + \cdot\cdot\cdot$
Keeping only the second-order approximation, we get
$\dfrac{1}{{\sqrt {1 - {k^2}{{\sin }^2}t} }} \approx 1 + \dfrac{1}{2}{k^2}{\sin ^2}t$
Recall that the period $T$ is given by
(1) ${\ \ \ \ \ }$ $T = 4\sqrt {\dfrac{L}{g}} E\left( k \right)$,
where $E\left( k \right) = \mathop \smallint \limits_0^{\dfrac{\pi }{2}} \dfrac{1}{{\sqrt {1 - {k^2}{{\sin }^2}t} }}{\rm{d}}t$ and $k = \sin \dfrac{\theta }{2}$.
For small $\theta $, we have $k = \sin \dfrac{\theta }{2} \approx \dfrac{\theta }{2}$.
Substituting $k \approx \dfrac{\theta }{2}$ and $\dfrac{1}{{\sqrt {1 - {k^2}{{\sin }^2}t} }} \approx 1 + \dfrac{1}{2}{k^2}{\sin ^2}t$ into equation (1) gives
$T \approx 4\sqrt {\dfrac{L}{g}} \mathop \smallint \limits_0^{\dfrac{\pi }{2}} \left( {1 + \dfrac{1}{8}{\theta ^2}{{\sin }^2}t} \right){\rm{d}}t$
Using $\cos 2x = 1 - 2{\sin ^2}x$ or ${\sin ^2}x = \dfrac{1}{2}\left( {1 - \cos 2x} \right)$, we get
$T \approx 4\sqrt {\dfrac{L}{g}} \mathop \smallint \limits_0^{\dfrac{\pi }{2}} \left( {1 + \dfrac{1}{{16}}{\theta ^2}\left( {1 - \cos 2t} \right)} \right){\rm{d}}t$
$T \approx 4\sqrt {\dfrac{L}{g}} \left[ {t + \dfrac{{{\theta ^2}}}{{16}}\left( {t - \dfrac{1}{2}\sin 2t} \right)} \right]_0^{\dfrac{\pi }{2}} = 4\sqrt {\dfrac{L}{g}} \left( {\dfrac{\pi }{2} + \dfrac{{{\theta ^2}}}{{16}}\cdot\dfrac{\pi }{2}} \right)$
Hence, $T \approx 2\pi \sqrt {\dfrac{L}{g}} \left( {1 + \dfrac{{{\theta ^2}}}{{16}}} \right)$.
