# Nasty Integral

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#1
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?
0
15 years ago
#2
(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.
0
15 years ago
#3
(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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