Jw L.

asked • 11/02/16

integration

integrate f(t) cos (n*pi*t)/T/2)dt with limit T/2 and 0

Arturo O.

What is f(t)?  This looks like a Fourier integral.
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11/02/16

Jw L.

f(t) = A
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11/02/16

1 Expert Answer

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Arturo O. answered • 11/03/16

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If f(t) = A = constant, then we have
 
A∫cos[(2πn/T)t]dt = AT/(2πn) sin[(2πn/T)t]
 
Evaluate a t = 0: 
 
AT/(2πn) sin[0] = 0
 
Evaluate at t = T/2:
 
AT/(2πn) sin[(2πn/T)(T/2)] = AT/(2πn) sin[πn] = 0
 
The definite integral just gives zero. 
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Jw L.

Why we need to times T/2(pi)n
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11/03/16

Arturo O.

Jw,
 
∫cos(kx)dx = (1/k) sin(kx)
 
You can verify this by differentiating (1/k) sin(kx).  You will get
 
(1/k) cos(kx) k = cos(kx)
 
In this problem, k = 2πn/T.  Therefore,
 
∫cos[(2πn/T)t] dt = [T/(2πn)] sin[(2πn/T)t]
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11/03/16

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