Dirac delta integral of cosx I have a pboblem as: $\int_0^{2\pi } {\delta \left( {{\rm{cos}}x} \right)dx} $.
I have done this: 
$\begin{array}{l}
g\left( x \right) = {\rm{cos}}x = 0 \Rightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2}\\
x =  - \frac{\pi }{2}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
g'\left( {\frac{\pi }{2}} \right) =  - \sin \left( {\frac{\pi }{2}} \right) =  - 1\\
g'\left( { - \frac{\pi }{2}} \right) =  - \sin \left( { - \frac{\pi }{2}} \right) = 1
\end{array} \right.\\
 \Rightarrow \int_0^{2\pi } {\delta \left( {{\rm{cos}}x} \right)dx}  = \int_0^{2\pi } {\delta \left( {x - \frac{\pi }{2}} \right)dx}  + \int_0^{2\pi } {\delta \left( {x + \frac{\pi }{2}} \right)dx = 2 } 
\end{array}$
Howerver, my pro said that I did it wrong, and require me rethouht!!!
Where is my false? Thanks
 A: $\newcommand{\+}{^{\dagger}}%
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$$
\color{#00f}{\large%
\int_{0}^{2\pi}\delta\pars{\vphantom{\LARGE A}\cos\pars{x}}\,\dd x}
=
\int_{0}^{2\pi}
\bracks{{\delta\pars{x - \pi/2} \over \verts{-\sin\pars{\pi/2}}}
+
{\delta\pars{x - 3\pi/2} \over \verts{-\sin\pars{3\pi/2}}}}\,\dd x
=
\color{#00f}{\large 2}
$$
