Use FTC to show that integral of cos(kx)dx from neg pi to pi = 0?

Question

I wind up with sin(k*pi)/k - [sin(-k*pi)/k]. I'm not sure if I'm right so far. But if I am I don't see how this =0.

Answer 1

I = ∫ cos (kx) dx------------ between given limits from = - π to π

I = (1/k) [ sin (kx) ]--------- between given limits

I = (1/k) [sin (kπ) - sin (- kπ ]

I = (1/k) [ sin(kπ) - sin(kπ) ]---(note that sin(-x) = -sin x)

I = 1/k x 0 = 0

Answer 2

yes your are right but sin(pi)=0 and also sin(Kpi)=0 or the negative is 0 all the time. The value of K doesn't really matter because it is a constant. It only has to be different from 0. So, the answer is 0

Answer 3

It's OK. And for every integer k, sin(k* pi) and sin()-k*pi) = 0. If k is even, then k*pi has the same determination as 0, so that sin(k*pi) = sin (0) = 0. And if k is odd, then k*pi has the same determination as pi, so that sin(k*pi) = sin (pi) = 0