#int\ e^(-7x)*cos(2x)dx =# ?

2 Answers
Feb 19, 2018

#int e^(-7x) cos 2x dx = (e^(-7x)(2sin2x-7cos2x))/53+C#

Explanation:

Integrate by parts:

#int e^(-7x) cos 2x dx = -1/7 int cos2x d(e^-7x) = -1/7 e^(-7x)cos2x - 2/7 int e^(-7x)sin2x dx#

and then again:

#int e^(-7x) cos 2x dx = -1/7 e^(-7x)cos2x + 2/49 e^(-7x)sin2x -4/49 int e^(-7x) cos 2xdx#

the same integral is now on both sides of the equation so we can solve for it:

#(1+4/49) int e^(-7x) cos 2x dx = -1/7 e^(-7x)cos2x + 2/49 e^(-7x)sin2x +C#

#int e^(-7x) cos 2x dx = (e^(-7x)(2sin2x-7cos2x))/53+C#

Feb 19, 2018

See below.

Explanation:

Using de Moivre's identity

#e^(ix) = cosx+i sinx# we have

#int\ e^(-7x)*cos(2x)dx = "Re"[int\ e^(-7x)e^(2i x) dx] = "Re"[int\ e^(2 ix-7x) dx]#

but

#int\ e^( 2i x-7x) dx =e^(i 2 x-7x)/ (2i-7)= -(i2+7)/53 e^(2i x-7x)=#

#= -e^(7x)(i2+7)/53(cos(2x)+isin(2x)) = e^(7x)/53(-7cos(2x)+2sin(2x))-i/53(2cos(2x)+7sin(2x))#

and finally

#int\ e^(-7x)*cos(2x)dx =e^(7x)/53(2sin(2x)-7cos(2x))+C_0#