Int. by parts please (1 Viewer)

[currysauce]

Hey anyone...

INT (e^x cosx) dx thanks

oh and... a subsitution one


use u = t - t^-1 to solve

INT ( ( 1+t^2) / (1+ t^4) )

cheers

 

[pLuvia]

∫e^xcosxdx

Let u=e^x dv=cosxdx
du=e^xdx v=sinx
I=e^xsinx-∫sinxe^xdx
From ∫sinxe^xdx
u=e^x dv=sinxdx
du=e^xdx v=-cosx
I₂=-e^xcosx+∫e^xcosxdx
=-e^xcosx+I

I=e^xsinx-[-e^xcosx+I]
2I=e^x(sinx+cosx)
I=[e^x(sinx+cosx)]/2

 

[_ShiFTy_]

I = INT (e^x cosx) dx

Integration by parts needs to be done twice in this question

u = e^x dv = cosx.dx
du = e^x.dx v = sinx.dx

e^x sinx - INT (e^x sinx) dx

INT (e^x sinx) dx

u = e^x dv = sinx.dx
du = e^x.dx v = -cosx.dx

-e^x cosx + INT (e^x cosx) dx
-e^x cosx + I

Combine them:

I = e^x sinx - [-e^x cosx + I]
2I = e^x (sinx + cosx)
I = e^x.(sinx + cosx)/2

 

[Trev]

int. [(1+t²)/(1+t⁴)] . dt
Taking a common factor of t² out of both numerator and denominator:
int. [({1/t²}+1)/({1/t²}+t²)] . dt
u=t-(1/t)
du/dt=1+(1/t²)
∴ du=[1+(1/t²)].dt
Since u²=t²+(1/t²)-2; u²+2=t²+(1/t²)
Integral then becomes:
int. [1/(u²+2)].du
etc.

 

[currysauce]

could you finish that?

i completed it but am not getting their answer.. so i'm making some silly mistake as usual... thanks

 

[_ShiFTy_]

Following Trev's working out, you should end up with inverse tan

 

[currysauce]

ok well i'm missing something obvious LOL

 

[currysauce]

anyone.. please finish it so i get this crap

 

[pLuvia]

After all of Trev's working
int. [1/(u²+2)].du
=1/sqrt2[tan^-1u/sqrt2]+C
=1/sqrt2[tan^-1(t-t^-1)/sqrt2]+C

 

[Trev]

I forgot what the integral turns into exactly from where I got up to, I knew it was tan^-1(blah) but wasn't sure what to put in front... Too lazy to find a standard integral table.