# Fourier integrals

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

• 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. Apr 18, 2017 at 14:19
• The second equation is badly formatted version of the orthogonality relationship Apr 18, 2017 at 14:55
• 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 Apr 18, 2017 at 15:02
• @Triatticus $\delta_{mn}/(2T)$ would not be better?
– Jon
Apr 18, 2017 at 17:48
• 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 Apr 18, 2017 at 18:49

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.