Figuring Out a Substitution for an Integral Problem

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SUMMARY

The integral problem discussed involves the expression \(\int \frac{du}{\sqrt{u-u^2} \cdot (1+ub)}\). The solution provided by Mathematica yields \(\sqrt{\frac{4}{b+1}} \cdot \texttt{arctan} \left ( \sqrt{\frac{(b+1)u}{1-u}} \right )\). A suggested substitution is to complete the square in the square root, transforming \(u - u^2\) into \(\frac{1}{4} - (u - \frac{1}{2})^2\). This leads to the substitution \(v = u - \frac{1}{2}\), simplifying the integral to \(\int \frac{dv}{\sqrt{1/4 - v^2}((1 + \frac{1}{2}b) + bv)}\), with \(v = \frac{1}{2}\sin(\theta)\) proposed as a plausible substitution.

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Zorba
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I am trying to figure out which substitution to use to get this integral done:

[tex]\int \frac{du}{\sqrt{u-u^2} \cdot (1+ub)}[/tex]

When I plug it into Mathematica I get:

[tex]\sqrt{\frac{4}{b+1}} \cdot \texttt{arctan} \left ( \sqrt{\frac{(b+1)u}{1-u}} \right )[/tex]

Any ideas about a suitable substitution?
 
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I would complete the square in the square root: [itex]u- u^2= -(u^2- u+ 1/4- 1/4)= 1/4- (u- 1/2)^2[/itex].

Then let v= u- 1/2 so that u= v+ 1/2, 1+ ub= (1+ (1/2)b)+ bv.

Now the integral is
[tex]\int\frac{dv}{\sqrt{1/4- v^2}((1+ (1/2)b)+ bv)}[/tex]
and [itex]v= (1/2)sin(\theta)[/itex] looks like a plausible substitution.
 

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