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

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Let;

`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)`

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

Sources:

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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

`I=e^(-x)(1/2)sin(2x)-int(-e^(-x))(1/2)sin(2x)dx`

`=(e^(-x)sin(2x))/2+(1/2)inte^(-x)sin(2x)dx`

Again integrating by parts on RHS,thus

`2I=e^(-x)sin(2x)+e^(-x)(-cos(2x))/2-int(-e^(-x)((-cos(2x))/2)dx`

`=e^(-x)sin(2x)-(1/2)e^(-x)cos(2x)-(1/2)inte^(-x)cos(2x)dx`

`2I=(1/2)e^(-x)(2sin(2x)-cos(2x))-(1/2)I`

`(2I+(1/2)I)=(1/2)e^(-x)(2sin(2x)-cos(2x)`

`5I=e^(-x)(2sin(2x)-cos(2x))`

`I=(1/5)e^(-x)(2sin(2x)-cos(2x))+c`

Ans.