$\int\limits_0^{2\pi } {{{\cos }^5}x} dx$

Let $I = \int\limits_0^{2\pi } {{{\cos }^5}x} dx$ ,Let $f\left( x \right) = {\cos ^5}x$

Now, we have
$f\left( {2\pi - x} \right) = {\left( {\cos \left( {2\pi - x} \right)} \right)^5} = {\left( {\cos x} \right)^5} = {\cos ^5}x = f\left( x \right)$

$\left[ {\int\limits_0^{2a} {f\left( x \right)dx = } \left\{ \begin{array}{l}2\int\limits_0^a {f\left( x \right)dx,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,f\left( {2a - x} \right) = f\left( x \right)} \\0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,f\left( {\left( {2a} \right) - x} \right) = - f\left( x \right)\end{array} \right.} \right]$

$\Rightarrow$ $I = 2\int\limits_0^\pi {{{\cos }^5}xdx}$

Again , we have
$f\left( {\pi - x} \right) = {\left( {\cos \left( {\pi - x} \right)} \right)^5} = - {\cos ^5}x = - f\left( x \right)$

$\Rightarrow$ $2\int\limits_0^\pi {{{\cos }^5}xdx = 0}$

Hence, $\int\limits_0^{2\pi } {{{\cos }^5}xdx = 2\int\limits_0^\pi {{{\cos }^5}xdx = 2 \times 0 = 0} }$