I do not understand why for the two integrals below, the result is always 0 unless | k | = n , in which case the result is π / 2 ∫ 0 π d θ cos ⁡ k θ cos ⁡ n θ ∫ 0 π d θ sin ⁡ k θ sin ⁡ n θI have tried using trigonometric identities to get a general solution but I had no luck understanding the nature of the integrals. If anyone can point out any hints or patterns, it would be much appreciated.

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I do not understand why for the two integrals below, the result is always 0 unless       |    k      |    =  n , in which case the result is   π      /    2      ∫          0        π    d  θ  cos  ⁡  k  θ  cos  ⁡  n  θ          ∫          0        π    d  θ  sin  ⁡  k  θ  sin  ⁡  n  θI have tried using trigonometric identities to get a general solution but I had no luck understanding the nature of the integrals. If anyone can point out any hints or patterns, it would be much appreciated.

vard6vv · Accepted Answer

I will show you how to do the first one. The second one is analogous.Recall the trigonometric identity   cos  ⁡  (  k  θ  )  cos  ⁡  (  n  θ  )  =      1    2        [    cos    ⁡    (    k    −    n    )    θ    +    cos    ⁡    (    k    +    n    )    θ    ]  . Plugging this into the integral, you get      1    2        ∫    0    π        [    cos    ⁡    (    k    −    n    )    θ    +    cos    ⁡    (    k    +    n    )    θ    ]    d  θ  =      1    2                            (                              sin            ⁡            (            k            −            n            )            θ                                k            −            n                          +                              sin            ⁡            (            k            +            n            )            θ                                k            +            n                          )            |        0    π    =  0But, be careful that the above is valid only for       |    k      |    ≠  n , because otherwise you have 0 in the denominator and the expression is undefined. It is a fairly common mistake to forget this.For k=n, we have that   cos  ⁡  (  k  θ  )  cos  ⁡  (  n  θ  )  =      1    2        [    cos    ⁡    0    +    cos    ⁡    (    k    +    n    )    θ    ]    =      1    2        [    1    +    cos    ⁡    (    2    n    θ    )    ]  . Evaluating the integral of this function you will get   π      /    2.Similarly, for k=-n, we have that   cos  ⁡  (  k  θ  )  cos  ⁡  (  n  θ  )  =      1    2        [    1    +    cos    ⁡    (    2    n    θ    )    ]   and the integral of this function again evaluates to   π      /    2.

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