# Evaluate the expression 2/pi int_0^pi e^(-t/2)cos2nt dt   How does this relate to 2/pi int_0^pi e^(-t/2)sin2nt dt ?

Using Euler's formula

cos2nt + isin2nt = e^(i2nt)

Therefore

int_0^pi e^(-t/2)(cos2nt + isin2nt) dt = int_0^pi e^((i2n-1/2)t) dt

= (e^((i2n-1/2)t))/((i2n-1/2))|_0^(pi) = ((e^(-pi/2)-1))/((i2n-1/2))

Now, using integration by parts and letting u=sin2nt and v=-2e^(-t/2)

int_0^pi e^(-t/2)sin2nt dt = -2e^(-t/2)sin2nt |_0^pi + int_0^pi 4n e^(-t/2)cos2nt dt

Let int_0^pi e^(-t/2)cos2nt dt = I...

Using Euler's formula

cos2nt + isin2nt = e^(i2nt)

Therefore

int_0^pi e^(-t/2)(cos2nt + isin2nt) dt = int_0^pi e^((i2n-1/2)t) dt

= (e^((i2n-1/2)t))/((i2n-1/2))|_0^(pi) = ((e^(-pi/2)-1))/((i2n-1/2))

Now, using integration by parts and letting u=sin2nt and v=-2e^(-t/2)

int_0^pi e^(-t/2)sin2nt dt = -2e^(-t/2)sin2nt |_0^pi + int_0^pi 4n e^(-t/2)cos2nt dt

Let int_0^pi e^(-t/2)cos2nt dt = I implies int_0^pi e^(-t/2)sin2ntdt = 4nI

We also have that int_0^pi e^(-t/2)(cos2nt+isin2nt) = ((e^(-pi/2)-1))/((i2n-1/2))= I + i4nI

Therefore I = ((e^(-pi/2)-1))/((i2n-1/2)(1+i4n)) = (2(1-e^(-pi/2)))/((1-i4n)(1+i4n)) = (2(1-e^(-pi/2)))/((1+16n^2))

and 4nI = (8n(1-e^(-pi/2)))/((1+16n^2))

Thus 2/piint_0^pi e^(-t/2)cos2nt dt = (4/pi)((1-e^(-pi/2)))/((1+16n^2)) = 0.504(2/((1+16n^2)))

and 2/pi int_0^pi e^(-t/2)sin2nt dt = 0.504((8n)/((1+16n^2))) answer

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You should use the formula of integration by parts such that:

int udv = uv - int vdu

u = cos(2nt) => du = -2n*sin(2nt)

dv = e^(-t/2)dt => v = int e^(-t/2) dt

Selecting -t/2 = x => -dt/2 = dx => dt = -2dx

int e^(t/2)dt = int e^x*(-2dx)

int e^x*(-2dx) = -2e^x

Substituting back -t/2  for x yields:

int e^(t/2)dt = -2e^(-t/2) => v = -2e^(-t/2)

You may evaluate the definite integral int_0^pi e^(-t/2)*cos(2nt)dt  such that:

int_0^pi e^(-t/2)*cos(2nt)dt = -2cos(2nt)*e^(-t/2)|_0^pi - int_0^pi (-2e^(-t/2))(-2n*sin(2nt)) dt

int_0^pi e^(-t/2)*cos(2nt)dt = -2cos(2nt)*e^(-t/2)|_0^pi - 4n int_0^pi (e^(-t/2))(sin(2nt)) dt

You need to use again integration by parts to evaluate int_0^pi (e^(-t/2))(sin(2nt)) dt  such that:

u = sin(2nt) => du = 2n*cos(2nt)

dv = e^(-t/2)dt => v = -2e^(-t/2)

int_0^pi (e^(-t/2))(sin(2nt)) = -2e^(-t/2)sin(2nt)|_0^pi+ 4n int_0^pi e^(-t/2) cos(2nt) dt

You should come up with the following notation for int_0^pi e^(-t/2)*cos(2nt)dt  such that:

int_0^pi e^(-t/2)*cos(2nt)dt = I

I= -2cos(2nt)*e^(-t/2)|_0^pi - 4n(-2e^(-t/2)sin(2nt)|_0^pi + 4nI)

I+ 16n^2I = -2cos(2nt)*e^(-t/2)|_0^(pi) + 8 n e^(-t/2)sin(2nt)|_0^pi

Factoring out I to the left side yields:

I(1 + 16n^2) = -2cos(2npi)*e^(-pi/2) + 2cos(2n0)*e^(-0/2) + 8 n e^(-pi/2)sin(2npi) - 8 n e^(-0/2)sin(2n*0)

I(1 + 16n^2) = -2e^(-pi/2) + 2 + 0 - 0

I = 2(1 - e^(-pi/2))/(1 + 16n^2)

Multiplying by 2/pi  yields:

2/pi*I = (4/pi)(1 - e^(-pi/2))/(1 + 16n^2)

Hence, evaluating the integral 2/pi*int_0^pi e^(-t/2)*cos(2nt)dt  yields 2/pi*int_0^pi e^(-t/2)*cos(2nt)dt = (4/pi)(1 - e^(-pi/2))/(1 + 16n^2).

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