I have to evaluate an integral $\int_{- \infty }^{\infty} \frac{dt}{t^2} \delta (\cos t)$
The notion i have used that, we only get $\delta (\cos t) $ when t is equal to $\frac{\pi}{2} $. Therefore the answer should be $\frac{4}{\pi^2}$.
Is it the right way I'm thinking?
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{-\infty}^{\infty}{\delta\pars{\cos\pars{t}} \over t^{2}} \,\dd t} & = \int_{-\infty}^{\infty}\,\,\sum_{n = -\infty}^{\infty} {\delta\pars{t - \bracks{n + 1/2}\pi} \over \verts{-\sin\pars{t}}} {\dd t \over t^{2}} = \sum_{n = -\infty}^{\infty}{1 \over \bracks{\pars{n + 1/2}\pi}^{\,2}} \\[5mm] & = {4 \over \pi^{2}}\bracks{2\sum_{n = 0}^{\infty}{1 \over \pars{2n + 1}^{2}}} = {8 \over \pi^{2}}\bracks{\sum_{n = 1}^{\infty}{1 \over n^{2}} - \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}} \\[5mm] & = {8 \over \pi^{2}}\bracks{{3 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}}} = {8 \over \pi^{2}}\,{3 \over 4}\,{\pi^{2} \over 6} = \color{#f00}{1} \end{align}