fourier transform of cos(kx) using the formula given. i want to find the fourier transform of $$ f(x) = \cos (kx)$$
 using the fourier transformation formula $$f(k)={1\over  \sqrt(2\pi)}\int _{-\infty}^\infty (f(x) e^{ikx}dk$$
how can i do that??

As tired mentioned, a delta function, $\delta(x)$, is a distribution that satisfies
$$\int _{-\infty}^\infty\delta(x-a)f(x)=f(a)$$
Indeed the Fourier transform of the function $g(x)=1$ is a delta function.So we should Write 
$$\delta(\omega)={1\over  \sqrt{2\pi}}\int _{-\infty}^\infty e^{i\omega x}dx$$
$$\mathcal{F}[f(x)](\omega)={1\over  \sqrt{2\pi}}\int _{-\infty}^\infty \cos (kx)\,e^{i\omega x}dx={1\over  \sqrt{2\pi}}\int _{-\infty}^\infty \left(\frac{e^{ikx}+e^{-ikx}}{2}\right) e^{i\omega x}dx$$
we have
$$\mathcal{F}[f(x)](\omega)={1\over  2\sqrt{2\pi}}\int _{-\infty}^\infty \left(e^{ix(\omega +k)}+e^{ix(\omega -k)}\right) dx$$
$$\mathcal{F}[f(x)](\omega)={1\over  2}\delta(\omega+k)+\frac 12\delta(\omega-k)$$
