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Perplexing integration - sqrt(1+sinx)

  1. Nov 16, 2012 #1
    Hello,

    I am trying to integrate sqrt(1+sinx) dx but I keep getting 2sqrt(1-sinx) (having initially substituted u=sinx), which is wrong!! Could anyone please tell me what I am doing wrong?
     
  2. jcsd
  3. Nov 16, 2012 #2

    micromass

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    We can't really know what you did wrong unless you tell us what you did!! So, can you show us your work?
     
  4. Nov 16, 2012 #3
    As I wrote, I substituted u=sinx and got that the integral should be equal to 2sqrt(1-sinx). Which is wrong, according to Wolfram and a simple differentiation test. I am simply not sure where I faltered. May you please advise?
     
  5. Nov 16, 2012 #4

    dextercioby

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    You're starting in the wrong place. You should be using a different substitution. Try tan x/2 = u.
     
  6. Nov 16, 2012 #5
    I would have thought so, except that I was instructed to use that substitution. :s
    It should still be correct, granted the algebra is. Yet the answer is incorrect!
     
  7. Nov 16, 2012 #6

    D H

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    peripatein, you need to show us your work. We cannot read your mind.
     
  8. Nov 16, 2012 #7
    u=sinx, du=cosxdx
    Hence, int=[sqrt(1+u)/sqrt(1-u^2)] du=[1/sqrt(1-u)] du.
    Hence, int=2sqrt(1-sinx)
     
  9. Nov 16, 2012 #8

    D H

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    Try that last step again.
     
  10. Nov 16, 2012 #9
    What do you mean? Dividing by sqrt(1+u)?
     
  11. Nov 16, 2012 #10

    SammyS

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    Do the integration:

    [itex]\displaystyle \int \frac{1}{\sqrt{1-u}}\,du\ .[/itex]
     
  12. Nov 16, 2012 #11
    But isn't that 2sqrt(1-u)?
     
  13. Nov 16, 2012 #12

    SammyS

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    sign error.
     
  14. Nov 16, 2012 #13
  15. Nov 16, 2012 #14
    Wrong link. In any case, that is NOT the answer. Question is, why?
     
  16. Nov 16, 2012 #15

    D H

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    They're the same answer *if* [itex]\cos x=\sqrt{1-\sin^2 x}[/itex]. That's not always the case, but you implicitly assumed that this is an identity in your u-substitution.
     
  17. Nov 16, 2012 #16

    SammyS

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    What is the correct answer?

    You have to be careful about sign behavior of the functions you're working with.

    [itex]\displaystyle \cos(x)=\sqrt{1-\sin^2(x)}\ \ [/itex] only when cos(x) ≥ 0.

    Otherwise, [itex]\displaystyle \ \cos(x)=-\sqrt{1-\sin^2(x)\ .}[/itex]
     
  18. Nov 16, 2012 #17
    Okay, thanks!
    I am also trying to evaluate the improper integral 1/[(x-1)^1/3] between 0 and 4. May I do so thus?:

    { lim b->1- int 1/[(x-1)^1/3] dx between 0 and b } + { lim b->1+ int 1/[(x-1)^1/3] dx between b and 4 }?

    Is there a another, better way to evaluate it?
     
  19. Nov 16, 2012 #18

    SammyS

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    It's best to start a new thread for this.

    It's a completely different problem.
     
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