${\int }_0^{2\pi }\cos x.{\cos }^{2}2x.{\cos }^{3}3x{\cos }^{4}4x\mathrm{......}{\cos }^{2022}2022x.dx$ is equals to

$f\left(x\right)=\cos x.{\cos }^{2}2x.{\cos }^{3}3x{\cos }^{4}4x\mathrm{.........}{\cos }^{2022}2022x$

Since ${\cos }^{k}kx={\cos }^{k}k(2\pi -x),$ for all $k\in N,f\left(x\right)=f(2\pi -x)$

Therefore $I={\int }_0^{2\pi }f\left(x\right).dx=2{\int }_0^{\pi }f\left(x\right).dx$

Now ${\cos }^{k}k(\pi -x)=\{\begin{array}{cc}{\cos }^{k}kx& whenkiseven\\ -{\cos }^{k}kx& whenkisodd\end{array}$

So $f\left(x\right)=-f(\pi -x),$ hence ${\int }_0^{\pi }f\left(x\right).dx=0.$ 
So I = 0