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Trig sub and Integrals.

  1. Dec 6, 2006 #1
    Ok guys here we go. this isI jsut a suggested question from our teacher in prep for our final. it a integral one. trig substion the prob I am having is near the end.. all the substition works great was down to the last bit and for some reason I must be doing the back substition wrong. so here it is.

    orginal question

    Integral of Sqr(1-4x^2) dx

    I do the substion of x=1/2 sin @
    (@= theta)
    and do all that stuff.

    after I integrat i have 1/4 @ + 1/8 sin2@

    so i do the back sub for just @ i get arcsine 2x which is correct but some reason i don;t get the back end right...

    some how 1/8 sin2@ = 1/2*x*sqr(1-4x^2)

    so if you could fight out the why i am going wrong and screwing up could you tell me.
  2. jcsd
  3. Dec 7, 2006 #2


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    You've made a mistake in computing the integral of [itex] \cos^{2}\theta [/itex]

  4. Dec 7, 2006 #3
    whawt was I supposed do then.. as far as i know i should use the fact that cos @^2 =1/2(1+cos 2@)
    if i did that wrong then what is the correct way?
  5. Dec 7, 2006 #4


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    [tex]\int \sqrt{1- 4x^2}dx[/tex]
    Let [itex]sin(\theta)= 2x[/itex]. Then [itex]cos(\theta)d\theta= 2 dx[/itex] so that [itex]\frac{1}{2}cos(\theta)d\theta= xdx[/itex] and [itex]\sqrt{1- 4x^2}= \sqrt{1- sin^2(\theta)}= cos(\theta)d\theta[/itex].

    The integral becomes
    [tex]\int cos^2(\theta)d\theta[/tex]
    Yes, you are correct that [itex]cos^2(\theta)= \frac{1}{2}(1+cos(2\theta)[/itex] so that integral is
    [tex]\frac{1}{2}(1+ cos(2\theta)d\theta[/tex]
    [tex]= \frac{1}{2}(\theta+ \frac{1}{2} sin(2\theta)+ C[/itex]
    Since [itex]sin(\theta)= 2x[/itex],[itex]x= arcsin(2x)[/itex] and
    [itex]sin(2\theta)= 2sin(\theta)cos(\theta)= 2x\sqrt{1-4x^2}[/itex]
    ([itex]cos(\theta)= \sqrt{1- sin^2(\theta)}= \sqrt{1- 4x^2}[/itex])
    this is
    [tex]\frac{1}{2}(2arcsin(2x)+ x\sqrt{1-4x^2}+ C[/itex]
  6. Dec 7, 2006 #5
    ok just wanna m ake sure I get this. to solve the sin (2@) you used an Identity that sin 2@ = 2sin@cos@

    and you can find cos cause we know what sin is.
  7. Dec 8, 2006 #6


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    Yes, correct.

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