Let $I_n={\int }_0^{\pi /2} {\cos }^{n}x\cos nxdx$ then $I_n:I_{n+1}$ is equal to

We have,

$\begin{array}{l}{I}_n={\int }_0^{\pi /2} {\cos }^{n}x\cos nxdx\\ \therefore l_{n+1}-I_n={\int }_0^{\pi /2} \left({\cos }^{n+1}x\cos (n+1)x-{\cos }^{n}x\cos nx\right)\cdot dx\\ \Rightarrow I_{n+1}-I_n={\int }_0^{\pi /2} {\cos }^{n}x\{\cos (n+1)x\cos x-\cos nx\}dx\\ \Rightarrow I_{n+1}-I_n={\int }_0^{\pi /2} {\cos }^{n}x{\{}_{ }^{ }\cos (n+1)x\cos x-\cos \left\{{\left((n+1)x-x\right)}_{ }^{ }\right\}dx\\ \Rightarrow I_{n+1}-I_n={\int }_0^{\pi /2} {\cos }^{n}x\{-\sin (n+1)x\sin x\}dx\\ \Rightarrow I_{n+1}-I_n={\int }_0^{\pi /2} \sin (n+1)x{\cos }^{n}x(-\sin x)dx\\ \Rightarrow I_{n+1}-I_n={\left[\sin (n+1)x\frac{{\cos }^{n+1}x}{n+1}\right]}_0^{\pi /2}-{\int }_0^{\pi /2} \cos (n+1)x{\cos }^{n+1}xdx\\ \Rightarrow I_{n+1}-I_n=-I_{n+1}\end{array}$

$\Rightarrow 2I_{n+1}=I_n\Rightarrow I_n:I_{n+1}=2:1$