Let $${a_n} = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x.\cos \,nx\,dx.} $$      Then $${a_n}:{a_{n + 1}}$$   is equal to :

Solution :
$$\eqalign{ & {a_{n + 1}} - {a_n} = \int_0^{\frac{\pi }{2}} {\left\{ {{{\cos }^{n + 1}}x.\cos \left( {n + 1} \right)x - {{\cos }^n}x.\cos \,nx} \right\}dx} \cr & = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x.\left\{ {\cos \,x.\cos \left( {n + 1} \right)x - \cos \,nx} \right\}} dx \cr & = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x.\left\{ { - \sin \,x.\sin \left( {n + 1} \right)x} \right\}dx} \cr & = \left[ {\sin \left( {n + 1} \right)x.\frac{{{{\cos }^{n + 1}}x}}{{n + 1}}} \right]_0^{\frac{\pi }{2}} - \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{n + 1}}x}}{{n + 1}}.\left( {n + 1} \right)\cos \left( {n + 1} \right)x\,dx} \cr & = - {a_{n + 1}} \cr & \therefore 2{a_{n + 1}} = {a_n} \cr} $$