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How do you integrate #e^(-x) * cos(2x) dx# ?

Answer 1

Scarlett Allison

The answer is #=e^-x/5(2sin(2x)-cos(2x))+C#

Perform integration by parts #2# times

#intuv'=uv-intu'v#

#u=cos2x#, #=>#, #u'=-2sin2x#

#v'=e^-x#, #=>#, #v=-e^-x#

Therefore,

#inte^-xcos2xdx=-e^-xcos2x-int2e^-xsin2xdx#

#u=2sin2x#, #=>#, #u'=4cos2x#

#v'=e^-x#, #=>#, #v=-e^-x#

#int2e^-xsin2xdx=-2e^-xsin2x+int4e^-xcos2x#

So,

#inte^-xcos2xdx=-e^-xcos2x-(-2e^-xsin2x+int4e^-xcos2x)#

#=-e^-xcos2x+2e^-xsin2x-4inte^-xcos2x#

#5inte^-xcos2xd=-e^-xcos2x+2e^-xsin2x#

#inte^-xcos2xd=1/5(2e^-xsin2x-e^-xcos2x)+C#

#=e^-x/5(2sin2x-cos2x)+C#

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Answer 2

Charles Arocha

To integrate ( e^{-x} \cdot \cos(2x) , dx ), you can use integration by parts. The integral evaluates to:

[ \frac{e^{-x}(2\sin(2x) - \cos(2x))}{5} + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.