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Integration: multiple “u-substitutions”

Q:  ∫x³/√(4 – x2) dx


A:  Use u-substitution to start:

u = x^2

du = 2x dx  or (1/2)du = x dx

Now note: ∫x³/√(4 – x2) dx = ∫x*x2/√(4 – x2) dx

So, plugging in our u’s to replace x’s, we get:

x*x2/√(4 – x2) dx = (1/2)u/√(4 – u) du

[I try to use colors to show the substitutions… Is it helpful or just confusing??]

So, our new and improved problem is:

(1/2)∫u/√(4 – u) du

Now time for another substitution:

s = 4 – u  →  u = 4 – s
ds = -du  →  -ds = du

(1/2)∫ u / √(4 – u) du = -(1/2)∫ (4 – s) / √(s) ds

Now, for some house keeping and “light algebra”

-(1/2)∫ (4 – s) / √(s) ds = -(1/2)∫ 4/√(s) ds + (1/2) ∫ s/√(s) ds

= -21/√(s) ds + (1/2)√(s) ds

NOW INTEGRATE:

-2 (2) s^(1/2) + (1/2) (2/3) s^(3/2) + C

-4 s^(1/2) + (1/3) s^(3/2) + C

Plug back in u’s to then plug back in x’s and get:

-4(4 – x2)^(1/2) + (1/3)(4 – x2)^(3/2)+ C

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