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Hard Integration

  1. Sep 27, 2009 #1
    How would you integrate this. I tried all integration methods available in Calculus II to no avail.

    ∫√(1+sin x) dx
     
  2. jcsd
  3. Sep 27, 2009 #2
    Did you try u = sinx? That should give you something easy to work with.
     
  4. Sep 28, 2009 #3
    I tried but d(sin x)/dx is cos x. We're missing the cos(x) here. :)
     
  5. Sep 28, 2009 #4

    lanedance

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    is this indefinite or does it have limits? if there is limits you may be able to use the symmetry with cos to simplify things
     
  6. Sep 28, 2009 #5
    It has limits [0,∏]. I know the cosine is a symmetrical function, but could you please expand a little on your symmetry argument in solving this integral.
    thank you
     
  7. Sep 28, 2009 #6

    lanedance

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    well if you draw the sin and cos functions, you might be able to convince yourself that (could also do it by hand using pi/2 funtion shifts & negative angles):

    based on symmetry of the sin function around pi/2
    [tex] \int_0^{\pi} dx \frac{1}{\sqrt{1+sin(x)}} = 2 \int_0^{\pi/2} dx \frac{1}{\sqrt{1+sin(x)}}[/tex]

    and based on symmetry of the sin & cos functions on the interval [0,pi/2]
    [tex] \int_0^{\pi/2} dx \frac{1}{\sqrt{1+sin(x)}} = \int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}[/tex]

    so you could re-write the integral as
    [tex] \int_0^{\pi/2} dx (\frac{1}{\sqrt{1+sin(x)}} + \frac{1}{\sqrt{1+cos(x)}}) [/tex]

    unfortunately after all that, i'm not sure whether it helps, remember a similar looking integral and though it might be worth a go... sorry ;(
     
    Last edited: Sep 28, 2009
  8. Sep 28, 2009 #7

    lanedance

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    so if you bought up to here
    [tex] \int_0^{\pi} dx \frac{1}{\sqrt{1+sin(x)}} = 2 \int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}[/tex]

    how about using half angle
    [tex] cos(x) = 2cos^2(x/2)-1 [/tex]

    then
    [tex] \int_0^{\pi/2} dx \frac{1}{\sqrt{1+cos(x)}}= \int_0^{\pi/2} dx \frac{1}{\sqrt{2} cos(x/2)}[/tex]

    which is getting maybe a little more tractable...?
     
  9. Sep 28, 2009 #8
    I hadn't looked at it too closely at first, but this one is a little tricky where you need another subtle substitution:

    [tex]\int \sqrt{1 + sinx} ~dx[/tex]
    u = sinx → x = sin-1u

    du = cosx dx

    [tex]\frac{du}{\cos(\sin^{-1}u)} = dx[/tex]

    [tex]\int \frac{\sqrt{1 + u}}{\cos(\sin^{-1}u)} du[/tex]
     
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