Respuesta :

syed514
sin(kx)=0=>x=0,π,...
∫0πsin(kx)dx=−1/kcos(kx)|π0=−1/k(cos(kπ)−cos(k∗0))
if k is odd then,=−1/k(−1−1)=−1/k(−2)=2/k
if k is even then,=−1/k(1−1)=0

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