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Int. by parts please (1 Viewer)

currysauce

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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
 
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pLuvia

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∫excosxdx

Let u=ex dv=cosxdx
du=exdx v=sinx
I=exsinx-∫sinxexdx
From ∫sinxexdx
u=ex dv=sinxdx
du=exdx v=-cosx
I2=-excosx+∫excosxdx
=-excosx+I

I=exsinx-[-excosx+I]
2I=ex(sinx+cosx)
I=[ex(sinx+cosx)]/2
 
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_ShiFTy_

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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
 
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Trev

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int. [(1+t²)/(1+t<sup>4</sup>)] . 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)
&there4; 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

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could you finish that?

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

_ShiFTy_

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

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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

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

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