Using the method of integration by parts, find ∫ e^−x  cos2xdx

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

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`U = e^(-x)`

`V = 1/2sin(2x)`


`dU = -e^(-x)dx`

`dV = cos(2x)dx`


`I = inte^-xcos(2x)dx`

By integral by parts;

`intUdV = UV-intVdU`


`inte^-xcos(2x)dx = (e^(-x)xx1/2sin2x)-int1/2sin(2x)(-e^(-x))dx`

`I = (e^(-x)xx1/2sin2x)+1/2intsin(2x)e^(-x)dx`


`I_1 = intsin(2x)e^(-x)dx`

By using integral by parts to the above equation;

`I_1 = e^(-x)(-1/2cos2x)-int(-1/2cos2x)(-e^(-x))dx`

`I_1 = e^(-x)(-1/2cos2x)-1/2I`


`I = (e^(-x)xx1/2sin2x)+1/2intsin(2x)e^(-x)dx`

`I = (e^(-x)xx1/2sin2x)+1/2I_1`

`I = (e^(-x)xx1/2sin2x)+1/2(e^(-x)(-1/2cos2x)-1/2I)`

`5I/4 = (e^(-x)xx1/2sin2x)+1/2e^(-x)(-1/2cos2x)`

`I = 4/5((e^(-x)xx1/2sin2x)+1/4e^(-x)cos2x)`

`I = 1/5e^-(x)(2sin2x+cos2x)`


So the answer is;

`I = 1/5e^-(x)(2sin2x+cos2x)`


Top Answer

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

We have given

`I=inte^(-x)cos(2x)dx`  ,   apply ILATE

so `e^(-x)`  is first function and cos(2x) is second function,thus integrating by parts, we have



Again integrating by parts on RHS,thus








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