# 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` ?

mathsworkmusic | (Level 2) Educator

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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

Sources:

sciencesolve | Teacher | (Level 3) Educator Emeritus

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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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