4

You can rewrite the integral as $$ \int_0^{2\pi} \left(\sum_{j=0}^{n-1}\cos\Big(\sin t+\tfrac tn+2\pi \tfrac jn\Big)\right)\,dt. $$ But $\sum_{j=0}^{n-1}\cos\big(a+2\pi \tfrac jn\big)=0$ for all $a$. In particular, the equality holds if $\sin t$ is replaced by any $2\pi$-periodic function.


Only top voted, non community-wiki answers of a minimum length are eligible