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Question
Hi,

I am stuck on the following integral:

sqrt(a - bx^2)dx

Actually, i need to solve the following integral for one of my physics problem but can't get my head around it:

integrate [sqrt(2*m*E - (1/2*m*omega^2*x^2)] dx

please help!!

Regards,
Puneet

Answer
Ok, the answer is :
∫√(a-bx^2)dx=(x/2)√(a-bx^2)+(√a/√b)arcsin[(√a/√b)x].
Where " √ " is the square root.
I will prove it to you in a very beautiful way :
1/√(a-bx^2)=(a-bx^2)/√(a-bx^2)=(2a-2bx^2)/2√(a-bx^2)=
(a-bx^2-bx^2+a)/2√(a-bx^2).
=(1/2)√(a-bx^2)-bx^2/2√(a-bx^2)+a/2√(a-bx^2)
=(1/2)√(a-bx^2)-(x/2)*bx/√(a-bx^2)+a/2√(a-bx^2)
=[ (x/2)√(a-bx^2) ]' + a/2√(a-bx^2).
Now lets concentrate on the last term :
a/2√(a-bx^2)=(a/2)*1/√a√[1-(b/a)x^2]=(a/2)*1/√a√[1-(x√b/√a)^2]
={√a/√b}*{√b/√a)/√[1-(x√b/√a)^2]
={√a/√b}* {arcsin[x√b/√a]}'
So, 1/√(a-bx^2)= { (x/2)√(a-bx^2) }' + √a/√b * { arcsin[x√b/√a] }'.
Hence,
∫√(a-bx^2)dx=∫ { (x/2)√(a-bx^2) }' + √a/√b * { arcsin[x√b/√a] }' dx
=(x/2)√(a-bx^2)+(√a/√b)arcsin[(√a/√b)x].

Alon.  

Calculus

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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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1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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M.A in Mathematics & Bs.c in Electronics.

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