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Calculus/Integration

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Expert: - 10/31/2008


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
Factor out √b, so the problem reduces to √b√(a/b - x²).  Let the constant a/b be c².  

Since we have the value √(c²-x²), lets use the triangular substitution with Θ the angle of the right triangle that is close to the side that has length √(c²-x²) and x to be the far side with c as the hypoteneuse.

sin(Θ)=x/c.  Differentiating gives cos(Θ)dΘ = 1/c dx.

⌠
⌡√(c²-x²)dx * √b, which converts by

c²-x² = c cos(Θ) and dx = c sec²(Θ) dΘ.

This converts our integral to

⌠
⌡c² cos(Θ) sec²(Θ) dΘ * √b.

Since we know that sec(Θ) is 1/cos(Θ), this quickly reduces to

⌠
⌡c² sec(Θ) dΘ * √b.

The value of this integral is c²ln(sec(Θ) + tan(Θ))√b.

Referring back to the problem, c² was a/b and putting in what the sec(Θ) and tan(Θ) are gives
(a/b) * ln(c/√(c²-x²) + x/√(c²-x²))
which can be reduced to (a/b) * ln((c+x)/√(c²-x²)).

The values are a = 2mE, b=mΩ²/2, and c = √(4E/Ω²).

Putting these in gives

(4E/Ω²)ln[(√{4E/Ω²} + x)/(4E/Ω² - x²)](mΩ²/2)

which simplifies to 2Em ln[1/(√{4E/Ω²} - x)].

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