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Thread: Integrating correct substitution,,

  1. #1
    Member i_zz_y_ill's Avatar
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    Integrating correct substitution,,

    Is this the right answer? I don't think it is somehow!

    $\displaystyle \int e^x\sqrt{1-e^2x}dx$ $\displaystyle e^x=sinu$
    $\displaystyle =\int e^x\sqrt{cos^2u}du$ $\displaystyle e^2x=sin^2u$
    $\displaystyle =\int sinucosudx$ $\displaystyle x=lnsinu$
    $\displaystyle dx=\frac{cosu}{sinu}du$
    $\displaystyle =\int cos^2udu$
    $\displaystyle =o.5\int cos2u+1du=$$\displaystyle \frac{sin2u}{4}+\frac{u}{2}+C$$\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C$
    Last edited by i_zz_y_ill; Nov 5th 2008 at 11:21 AM.
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  2. #2
    Moo
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    Quote Originally Posted by i_zz_y_ill View Post
    Is this the right answer? I don't think it is somehow!

    $\displaystyle \int e^x\sqrt{1-e^2x}dx$
    I used $\displaystyle e^x=sinu$
    =\int e^x\sqrt{cos^2u}
    =\int sinu cosudx
    =\int cos^2udu
    =o.5\int cos2u+1du
    =\frac{sin2u}{4}+\frac{u}{2}+C
    =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C[/tex]
    Hello,

    I didn't read 'til the end because there are problems with the latex ><

    the substitution is okay.

    $\displaystyle e^x=\sin(u) \implies e^x dx=\cos(u) du \implies dx=\frac{\cos(u)}{e^x} ~ dx$
    so the e^x will simplifiy with the one that is already in the integral.




    PS : for the latex code, leave a space after \int, otherwise, they'll understand \intsinu as a single command, which doesn't exist
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  3. #3
    Member i_zz_y_ill's Avatar
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    I did along the lines f what you said,my original post now has my working,,i think its wrong though!!! and can i not just substitute u^2=1-e^2x thats where im a bit confused i get a very simple answer when i do compared to my other method?
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  4. #4
    Moo
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    It's still very confusing

    $\displaystyle e^x=\sin(u)$

    Then $\displaystyle \dots=\int \cos(u) \sin(u) ~ du$

    now, just substitute $\displaystyle y=\sin(u)$


    and can i not just substitute u^2=1-e^2x
    Yes you can
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  5. #5
    Member i_zz_y_ill's Avatar
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    hmm i never got $\displaystyle \int sinucosudu$ i got $\displaystyle \int sinucosudx$ I think if you look at my first post i have the answer ,, i used $\displaystyle e^x=sinu$ as the substitution except the answer looks very messy''.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    $\displaystyle \int{e^x}\sqrt{1-e^{2x}}dx\overbrace{\mapsto}^{u=e^x}\int\sqrt{1-u^2}du\overbrace{\mapsto}^{u=\sin(z)}\int\cos^2(z) dz$
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  7. #7
    Member i_zz_y_ill's Avatar
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    oh right thanks,,,so that gives me
    $\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C
    $ which i cant simplify?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    oh right thanks,,,so that gives me
    $\displaystyle =\frac{sin(2sin^-1(e^x))}{4}+\frac{sin^-1(e^x)}{2}+C
    $ which i cant simplify?
    $\displaystyle \sin\left(2\arcsin\left(e^x\right)\right)=2\sin\le ft(\arcsin\left(e^x\right)\right)\cos\left(\arcsin \left(e^x\right)\right)=2e^x\sqrt{1-e^{2x}}$
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  9. #9
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    I see where you get $\displaystyle 2sinarcsin(e^x)cos(arcsin(e^x)$ but I cannot see how you got to the next line,,,,and does the term $\displaystyle \frac{arcsin(e^x)}{2}$ also simplify?? thanks I really appreciate this
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by oxrigby View Post
    I see where you get $\displaystyle 2sinarcsin(e^x)cos(arcsin(e^x)$ but I cannot see how you got to the next line,,,,and does the term $\displaystyle \frac{arcsin(e^x)}{2}$ also simplify?? thanks I really appreciate this
    $\displaystyle \sin\left(\arcsin(f(x))\right)=f(x)$

    $\displaystyle \cos\left(\arcsin(f(x))\right)=\sqrt{1-f(x)^2}$

    And no, not to anything you would want.
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  11. #11
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    how is $\displaystyle cos(arcsin(e^x)=\sqrt{1-exp(2x)}$?i cant see this
    Last edited by oxrigby; Nov 5th 2008 at 01:47 PM.
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