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    As the tree said to the lumberjack, I'm stupid... I mean stumped

    Anyway, I'm trying to integrate

    (1-x^2)^(3/2) [wrt x]

    I just don't really know where to start - is there a substitution or another way of writing it that makes it easier?
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    (Original post by loismustdie)
    As the tree said to the lumberjack, I'm stupid... I mean stumped

    Anyway, I'm trying to integrate

    (1-x^2)^(3/2) [wrt x]

    I just don't really know where to start - is there a substitution or another way of writing it that makes it easier?
    If x=sinu
    dx/du = cosu
    dx = cosu du

    (1-x^2)^(3/2) dx
    = (cos^2 u)^(3/2) . cosu du
    = cos^4 u du
    cos2u = 2cos^2 u -1
    cos^2 u = 0.5(1+cos2u)
    = 0.25[1+cos2u]^2 du
    = 0.25[1 + 2cos2u + cos^2 2u] du
    = 0.25[1 + 2cos2u + 0.5 + 0.5cos4u] du
    = 0.25[1.5 + 2cos2u + 0.5cos4u] du
    = 0.25[1.5u + sin2u + (1/8)sin4u] +C
    You can then rewrite that in terms of sinu and hence return it back into x.
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    (Original post by loismustdie)
    Find Int. (1 - x^2)^(3/2) dx
    Let x = sinu --> dx/du = cosu --> dx = cosu du
    Also note: u = sin^-1(x)

    Hence: Int. (1 - x^2)^(3/2) dx.
    = Int. (1 - sin^2u)^(3/2).cosu du.
    = Int. (cos^2u)^(3/2).cosu du.
    = Int. cos^4u du.
    = Int. cos^2u.cos^2u du.
    = Int. 1/2(cos2u + 1).1/2(cos2u + 1) du.
    = 1/4 Int. cos^2(2u) + 2cos2u + 1 du.
    = 1/4 Int. 1/2(cos4u + 1) + 2cos2u + 1 du.
    = 1/4 Int. (cos4u + 4cos2u + 3)/2 du.
    = 1/8 Int. (cos4u + 4cos2u + 3) du.
    = 1/8 [(sin4u/4 + 2sin2u + 3u] + k
    = sin4u/32 + sin2u/4 + 3u/8 + k
    = (sin4u + 8sin2u + 12u)/32 + k
    = {sin[4sin^-1(x)] + 8sin[2sin^-1(x)] + 12sin^-1(x)}/32 + k

    Nima
 
 
 
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