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Calculus/where is the error?
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Question
hi I am constantly trying to solve this problem WITHOUT TRIG SUBSTITUTION
integral (9-x^2)^1/2 * (x-3)^2 dx
(3-x)^(1/2) *(x+3)^(1/2) *(x-3)^2 dx
(3-x)^5/2 (3+x) dx
u = 3-x
-u= x-3
-u+6= x+3
du/dx =-1
(u)^5/2 * (6-u) -du
-[6* u^(5/2) - u^7/2] du
-[12/7 u^(7/2) - u ^9/2 +C] du
-[12/7 (x-3)^(7/2) - (x-3)^9/2 +C]
but there is something wrong right?
Hello,
(3-x)^(1/2) *(x+3)^(1/2) *(x-3)^2 dx
(3-x)^5/2 (3+x) dx <--------- here is the problem . It should be (3+x)^(1/2)
u = 3-x
-u= x-3
-u+6= x+3
du/dx =-1
(u)^5/2 * (6-u) -du <--------- It should be (u)^5/2 * (6-u)^(1/2) -du
Ofcourse this integration is not immediate, it involves sinh(x) function .
Alon.
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Answers by Expert:
Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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