$$ \int_0^L (1-cos \frac{\pi x}{2l} )$$
$$ \int_0^l 1dx - \int_0^l cos \frac{\pi x}{2l} $$
$$ \int_0^l1 dx=x |^{l} _{0}=l-0=l $$
$$ \int_0^l cos \frac{\pi x}{2l} dx = \frac{1}{ \frac{\pi}{2l} } \int_0^l cos( \frac{\pi x}{2l} ) ( \frac{\pi}{2l} )= \frac{1}{ \frac{\pi}{2l} } sin( \frac{\pi x}{2l} ) |^{l} _{0} $$
$$ ( \frac{2l}{\pi} )sin( \frac{\pi x}{2l} ) |^{l} _{0} =( \frac{2l}{\pi} )(sin( \frac{\pi}{2} )-sin(0)) =( \frac{2l}{\pi} )$$
$$ \int_0^L (1-cos \frac{\pi x}{2l} )=l- \frac{2l}{\pi} $$
