How to integrate E^(-2x)(sin[x]) and E^(-2x)(cos[x])? Detail calculation need for further understanding.TQ?

Question

How to integrate E^(-2x)(sin[x]) and E^(-2x)(cos[x])? Detail calculation need for further understanding.TQ?

Answer 1

Integrate by parts twice. As part of the result you'll will get the original integrals. SOlve this equation for this integrals.

∫ e^(-2x) ∙ sin(x) dx

integrate by parts with

u' = sin(x) => u= -cos(x)

v = e^(-2x) => v' = -2e^(-2x)

∫ e^(-2x)∙sin(x) dx = -e^(-2x)∙cos(x) - ∫ 2∙e^(-2x)∙cos(x) dx  

<=>

∫ e^(-2x)∙sin(x) dx = -e^(-2x)∙cos(x) - 2∙ ∫ e^(-2x)∙cos(x) dx  

integrate by parts with

u' = cos(x) => u = sin(x)

v = e^(-2x) => v' = -2e^(-2x)

∫ e^(-2x)∙sin(x) dx = -e^(-2x)∙cos(x) - 2∙ ( e^(-2x)∙sin(x) + ∫2∙e^(-2x)∙sin(x) dx )

<=>

∫ e^(-2x)∙sin(x) dx = -e^(-2x)∙cos(x) - 2∙e^(-2x)∙sin(x) - 4∙∫ e^(-2x)∙sin(x)  dx

solve for ∫ e^(-2x)∙sin(x)  dx

∫ e^(-2x)∙sin(x) dx = -(1/5)∙e^(-2x)∙( cos(x) +∙ 2∙sin(x) )

Same procedure for the other integral:

∫ e^(-2x)∙cos(x) dx = e^(-2x)∙sin(x) + 2∙ ∫ e^(-2x)∙sin(x) dx  

<=>

∫ e^(-2x)∙cos(x) dx = e^(-2x)∙sin(x) + 2∙( -e^(-2x)∙cos(x) - ∫ 2∙e^(-2x)∙cos(x) dx )

<=>

∫ e^(-2x)∙cos(x) dx = e^(-2x)∙sin(x) - 2∙e^(-2x)∙cos(x) - 4∙ ∫ e^(-2x)∙cos(x) dx )

<=>

∫ e^(-2x)∙cos(x) dx = (1/5)∙e^(-2x)∙( sin(x) - 2∙cos(x) )

Answer 2

Let c=cos(x), s=sin(x) for convenience!

♠ y*dx =y1*dx +j*y2*dx, where

y1 = exp(-2x) *c, y2=exp(-2x) *s;

thus y*dx = dx*exp(-2x) *(c +j*s) =

=dx*exp(-2x)*exp(jx) = dx*exp((-2 +j)*x);

♣ hence Y= exp((-2 +j)*x) /(-2 +j) =

= [(-2 -j)/(4+1)] *exp(-2x)*(c +j*s) =

=0.2* exp(-2x)*(-2*c +s –j*c –2j*s) =

= 0.2* exp(-2x)*(-2*c +s) +0.2j* exp(-2x)*(-c -2s);

♦ therefore ∫ dx*exp(-2x) *cos x =0.2* exp(-2x)*(-2*c +s) +C;

♦ therefore ∫ dx*exp(-2x) *sin x =0.2* exp(-2x)*(-c -2s) +C;