Question

Integrate e^xcosx dx by parts?

Answer 1

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

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

Explanation:

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`