Something needs to be reduced (2)

The problem can be solved in two cases

Case(1) When $n$ is odd

Use the rule $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$

Here the cosine part will give negative and also it is raised i odd powers for all $n$ .And sine part will give positive.So, for every odd $n$ the value of integral will be zero.

Case(2) When $n$ is even

Again use the rule $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$

Here every sine part will give a negative sign and also it is raised in odd powers and cosine part will give positive.So product is negative and value of integral will be zero for all even $n$

So, total sum will be $0$