when they took integral of both sides then $\int \cos nwt \cos mwt dt$ will be equal to $0$ but $a_n$ stays but $\Sigma$ or sum of it will be neglected so i don't understand why. We still have our $a_n$ with $n$. why do we have to get rid of it when we still have same coefficients on the formula?
Remember that
$$ \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y $$
Therefore
\begin{eqnarray} \cos (m + n)\omega t &=& \cos m\omega t \cos n \omega t - \sin m\omega t \sin n \omega t \\ \cos (m - n)\omega t &=& \cos m\omega t \cos n \omega t + \sin m\omega t \sin n \omega t \end{eqnarray}
Adding both equations
$$ \cos m\omega t \cos n \omega t = \frac{1}{2}[\cos (m + n)\omega t + \cos (m - n)\omega t] $$
Consider now two cases
In this case both $m + n$ and $m - n$ are integers and
\begin{eqnarray} \int_0^T {\rm d}t~ \cos m\omega t \cos n \omega t &=& \frac{1}{2} \int_0^T {\rm d}t~\cos (m + n)\omega t + \frac{1}{2} \int_0^T {\rm d}t~\cos (m - n)\omega t \\ &=& \left.\frac{1}{2\omega(m + n)}\sin (m + n)\omega t \right|_0^T + \left.\frac{1}{2\omega(m - n)}\sin (m - n) \omega t \right|_0^T \\ &=& \frac{\sin 2\pi (m + n)}{2\omega(m + n)} + \frac{\sin 2\pi (m - n)}{2\omega(m - n)} = 0 \end{eqnarray}
where I have used the fact that $\sin 2\pi k = 0$ for $k$ integer
In this case
\begin{eqnarray} \int_0^T {\rm d}t~ \cos m\omega t \cos n \omega t &=& \frac{1}{2} \int_0^T {\rm d}t~\cos 2n \omega t \\ &=& \left.\frac{1}{4n\omega} \sin 2n \omega t\right|_0^T \\ &=& \frac{\sin 4\pi n}{4n\omega} = 0 \end{eqnarray}