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Finding integral

  • Thread starter Doc G
  • Start date
  • #1
18
0
Hi,

Homework Statement


find value of
[tex]
\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx
[/tex]

Homework Equations



The Attempt at a Solution



I first used subsitution u = lnx, that gave me:

[tex]
\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]
then letting
[tex]
I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]
integrating by parts twice, gave me:
[tex]
I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
[/tex]
[tex]
I = - e^{u} \cos u + e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]

therefore

[tex]
I = - e^{u} \cos u + e^{u} \cos u + I
[/tex]
which of course cancels to zero which is not very useful

Any ideas on finding the value of this integral, or have i made an error in my working?

Many Thanks
 
Last edited:

Answers and Replies

  • #2
2,063
2
I think you need to re-check the 'by-parts'.
 
Last edited:
  • #3
CompuChip
Science Advisor
Homework Helper
4,302
47
Yep. In particular, in the second "by parts", think where that sine inside the integral comes from :)
 
  • #4
18
0
Taking the result of the first integration by parts:

[tex]
I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
[/tex]

then working with the integral part:
[tex]
+\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
[/tex]
let u= cos u
therefore
[tex]
\frac{du}{dx}= - sinu
[/tex]
let
[tex]
\frac{dv}{dx}= u e^{u}
[/tex]
therefore
[tex]
v = e^{u}
[/tex]

then using identity that:

[tex]
\int u \frac{dv}{dx} \equiv uv - \int v \frac{du}{dx}
[/tex]

I got
[tex]
e^{u} \cos u + \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]

Note- i chose u=cos u and dv/dx =ue^u [rather than u=ue^u]since i thought otherwise I would never be able to achieve an equivalent expression without an integral involved as I would get greater and greater powers of u in front of e^u as I kept going.

Sorry if i'm not seeing something obvious, but i cant find the error there

thanks for the replies
 
Last edited:
  • #5
2,063
2
Note- i chose u=cos u and dv/dx =ue^u [rather than u=ue^u]since i thought otherwise, I thought I would never be able to achieve an equivalent expression without an integral involved as I would get great and great powers of u in front of e^u as I kept going.
How did you get a 'u' within the integral at the result of the first integration by parts? (And it's now the derivative with respect to u, not x)

[I'm using t instead of u for the sake of clarity]

The first integral, after substitution [tex]= \int_{0}^{\pi/2}{e^t\sin{t}}dt[/tex]

Let u = cos(t) and dv/dt = e^t. Therefore, du/dt = -sin(t) and v = e^t.
 
Last edited:
  • #6
18
0
Thanks, I think i've got it now; but just to check here my full working:

let
[tex]
I = \int_{0}^{\frac{\pi}{2}} e^{t}\sin t dt
[/tex]

u = sin t
dv/dt= e^t

therefore
du/dt = cos t
v = e^t

so:
[tex]
I = e^{t} sin t - \int_{0}^{\frac{\pi}{2}} e^{t} cos t dt
[/tex]

and integrating by parts again:
u = cos t
dv/dt = e^t

so
du/dt = -sin t
v = e^t

[tex]
I = e^{t} sin t - e^{t} cos t - \int_{0}^{\frac{\pi}{2}} e^{t} sin t dt
[/tex]

[tex]
I = e^{t} sin t - e^{t} cos t - I
[/tex]

[tex]
2I = [e^{t} ( sin t - cos t )]_{0}^{\frac{\pi}{2}}
[/tex]
[tex]
I = \frac{1}{2} [(e^{\frac{\pi}{2}}( 1 - 0 )) - ( -1)]
[/tex]

and finally, giving
[tex]
I = \frac{1}{2} [e^{\frac{\pi}{2}} + 1 ]
[/tex]

Hopefully its right this time?

Thanks again
 
Last edited:
  • #7
2,063
2
Perfect! ...
 
  • #8
18
0
Great! Thank you very much for the advice
 
  • #9
2,063
2
You're welcome. :)
 
  • #10
CompuChip
Science Advisor
Homework Helper
4,302
47
Hmm, I don't quite follow what you're doing in your subsequent posts (it looks correct though), but your first try was already correct if it weren't for one minus sign:

[tex]
\int_{1}^{e^\frac{\pi}{2}} \sin (lnx) dx
[/tex]

I first used subsitution u = lnx, that gave me:

[tex]
\int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]
then letting
[tex]
I = \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]
integrating by parts twice, gave me:
[tex]
I = - e^{u} \cos u +\int_{0}^{\frac{\pi}{2}} u e^{u}\cos u du
[/tex]
[tex]
I = - e^{u} \cos u + e^{u} \cos u - \int_{0}^{\frac{\pi}{2}} e^{u}\sin u du
[/tex]
Note the minus sign: cos' = -sin

therefore

[tex]
I = - e^{u} \cos u + e^{u} \cos u - I
[/tex]
so
[tex]2I = - e^{u} \cos u + e^{u} \cos u[/tex]
and the answer follows.
 
  • #11
2,063
2
Hmm, I don't quite follow what you're doing in your subsequent posts (it looks correct though), but your first try was already correct if it weren't for one minus sign:


so
[tex]2I = - e^{u} \cos u + e^{u} \cos u[/tex]
and the answer follows.
And you end up with 0, which is incorrect.
 

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