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Let `I=`` ` `int_0^pi (e^(-2x)) Cos x dx`   and `J=` `int_0^pi (e^-(2x)) sin x dx` .By...

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roshan-rox | Valedictorian

Posted June 30, 2013 at 3:08 PM via web

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Let `I=`` ` `int_0^pi (e^(-2x)) Cos x dx`   and `J=` `int_0^pi (e^-(2x)) sin x dx` .By using the metod of intergration by parts.Show that I=2J and J=1+e^-2`pi` -2I Hence obtain the values of I and J.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 30, 2013 at 3:44 PM (Answer #1)

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`I = int_0^pi(e^(-2x)cosx)dx`

`J = int_0^pi(e^(-2x)sinx)dx`

 

`I = int_0^pi(e^(-2x)cosx)dx`

`u = sinx`

`du = cosxdx`

`v = e^(-2x)`

`dv = -2e^(-2x)dx`

 

From inegral by parts

intvdu = uv-intudv

`int_0^pi(e^(-2x)cosx)dx = [sinxe^(-2x)]_0^pi+int_0^pi(sinx(-2e^(-2x)))dx`

`I = (0-0)+2int_0^pi(e^(-2x)sinx)dx`   (`sinpi `= `sin0`= 0)

` ` `I = 2J`

 

 

`J = int_0^pi(e^(-2x)sinx)dx`

`u = -cosx`

`du = sinxdx`

`v = e^(-2x)`

`dv = -2e^(-2x)`

 

By integral by parts

`int_0^pi(e^(-2x)sinx)dx = [-cosxe^(-2x)]_0^pi-int_0^pi(-cosx(-2e^(-2x)))dx`

`J = [-cospie^(-2pi)-(-cos0e^0)]-2int_0^pi(e^(-2x)cosx)dx`

`J = 1+e^(2pi)-2I`

 

We obtained that;

`I = 2J ---(1)`

`J = 1+e^(2pi)-2I`

From (1)

`J = 1+e^(2pi)-2xx2J`

`J = 1+e^(2pi)-4J`

`5J = 1+e^(2pi)`

`J = (1+e^(2pi))/5`

 

From (1)

`I = (2(1+e^(2pi)))/5`

 

So all the required answered are obtained.

`I = 2J`

`J = 1+e^(2pi)-2I`

`J = (1+e^(2pi))/5`

`I = (2(1+e^(2pi)))/5`

 

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