Consider the integral. In=∫cosnxdx The above integral is solved as, In=∫cosn−1xcosnxdx =cosn−1x‌sin‌x+∫cosn−2x(n−1)sin2xdx =cosn−1x‌sin‌x+(n−1)∫cosn−2x−(n−1)‌∫cosnxdx =cosn−1x‌sin‌x+(n−1)In−2−(n−1)In Solve further, In(1+n−1)=cosn−1x‌sin‌x+(n−1)In−2 nIn−(n−1)In−2=cosn−1x‌sin‌x Therefore, 6I6−5I4=cos5x‌sin‌x