Integrate e^xcosx dx by parts?

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Answer 1 · 2018-04-12T07:21:37

Let, #I=int e^x cosx dx #

Integrating by parts,

#int e^x cosx dx =I = int cosx d(e^x) = e^xcosx + int e^xsinx dx+C#

Again integrating #int e^xsinx dx# by parts:

#I = e^xcosx + e^xsinx - int e^xcosx dx+C#

#I = e^xcosx + e^xsinx - I+C#

#2I = e^xcosx + e^xsinx +C#

#I= (e^xcosx + e^xsinx)/2 +C#

#I= e^x/2(cosx + sinx) +C#

Answer 2 · 2018-04-12T07:49:00

#=>I=e^x/2(cosx+sinx)+c#

Here,

#color(blue)(I=inte^xcosx dx...to(A)#

#"Using"color(red)" Integration by Parts"#

i.e. #int(u*v)dx=uintvdx-int(u'intvdx)dx#

Let,, #u=cosxand v=e^x=>u'=-sinx and intvdx=e^x#

So,

#I=cosxe^x-int((-sinx)e^x)dx#

#=e^x*cosx+intsinxe^xdx#

Again #"using"color(red)" Integration by Parts"#

Take, #u=sinx and v=e^x=>u'=cosxand intvdx=e^x#

#:.I=e^xcosx+[sinxe^x-color(blue)(intcosxe^xdx)]+c#

#=>I=e^xcosx+sinxe^x-color(blue)I+c...toUse . (A) above#

#:.I+I=e^xcosx+e^xsinx+c#

#=>2I=e^x(cosx+sinx)+c#

#=>I=e^x/2(cosx+sinx)+c#