I'm starting with the fourier theme and I need to demonstrate a couple of integrals, but I'm coming back and I do not understand much, could you please help

  1. $\int_d^{d+T}\sin(n\omega t)dt = 0$

  2. $\int_d^{d+T}\sin(m\omega t)\sin(n\omega t)dt = \left\{ \begin{array}{ll} 0 \text{ for } m\neq n \\ \frac{1}{2T} \text{ for } m=m \end{array}\right.$

I would greatly appreciate your help

  • $\begingroup$ By sen, do you mean $\sin$? Also, I'm not sure what you're trying to say in number $2$ relating $m$ and $n$ and the results of the integral. $\endgroup$
    – user12345
    Apr 18, 2017 at 14:19
  • $\begingroup$ The second equation is badly formatted version of the orthogonality relationship $\endgroup$
    – Triatticus
    Apr 18, 2017 at 14:55
  • $\begingroup$ I've suggested an edit to improve the readability of your conditional equation and now that I think of it I put the not equals sign in the wrong place $\endgroup$
    – Triatticus
    Apr 18, 2017 at 15:02
  • $\begingroup$ @Triatticus $\delta_{mn}/(2T)$ would not be better? $\endgroup$
    – Jon
    Apr 18, 2017 at 17:48
  • $\begingroup$ Depends on how people feel about it, honestly it doesn't matter no? Also @user426995 you just need to re-edit to adjust where the not equal signs are, plus there is an m where there should be an n that was my mistake $\endgroup$
    – Triatticus
    Apr 18, 2017 at 18:49

2 Answers 2


Hint: complex definitions of sine makes two into an elementary integral. No trig identities required.


This a wonderful exercise that uses trig identities. I assume you can do the first integral. If you have any difficulty the answer is $\frac{-1}{nw} [cos(n w (d+T)) - cos(n w(d))]$The second one can be completed by using the trig identity $sin(x)^{2}= \frac{1}{2}(1 - cos(2x))$ for n=m case. Then you can use $sin(a)sin(b) = \frac{1}{2}[cos(a+b)-cos(a-b)]$ identity for the other. Hope this answers your question.


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