# Integrating the square root of a polynomial

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• Oct 7th 2009, 03:14 AM
The Power
Integrating the square root of a polynomial
How would I begin to solve this integrand ? by completing the square ?

I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.

$\displaystyle \int{\frac{1}{\sqrt{-4x^2+12x-4}}}$

This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.
• Oct 7th 2009, 03:34 AM
mr fantastic
Quote:

Originally Posted by The Power
How would I begin to solve this integrand ? by completing the square ?

I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.

$\displaystyle \int{\frac{1}{\sqrt{-4x^2+12x-4}}}$

This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.

Note that $\displaystyle -4x^2 + 12x - 4 = -4(x^2 - 3x + 1) = -4\left[\left( x - \frac{3}{2}\right)^2 - \frac{5}{4}\right] = 4 \left[\frac{5}{4} - \left( x - \frac{3}{2}\right)^2\right]$

$\displaystyle = 5 - (2x - 3)^2 = (\sqrt{5})^2 - (2x - 3)^2$.
• Oct 7th 2009, 03:48 AM
The Power
So, I was going in the general direction correctly by using completing the square method ?
• Oct 7th 2009, 04:43 AM
knivmannen
Hi,

I dont really know where fantastic is going with his tip but the problem looks like an arcsin/arccos to me,

the definition is:

derivative of arcsin(x) is:

derivative of arccos(x) is:

Make it look like 1 + something^2 in the square root and the solution should not be far away.
• Oct 7th 2009, 05:02 AM
tonio
Quote:

Originally Posted by knivmannen
Hi,

I dont really know where fantastic is going with his tip but the problem looks like an arcsin/arccos to me,

the definition is:

derivative of arcsin(x) is:

derivative of arccos(x) is:

Make it look like 1 + something^2 in the square root and the solution should not be far away.

I think you meant "...1 - something^2...", and this is what I believe fantastic was heading to: he already got that the expression within the square root is (sqrt(5))^2 - (2x - 3)^2 = 5 {1 -[(2x-3)/Sqrt(5)]^2}, and if we get this into the square root in the denominator this is exactly an Arcsin integral.

Tonio
• Oct 7th 2009, 10:22 AM
The Power
Quote:

Originally Posted by mr fantastic
[tex] = 5 - (2x - 3)^2

Where that last bit comes from ? I see where 5 comes from, but not
(2x - 3)^2
• Oct 7th 2009, 10:46 AM
tonio
Quote:

Originally Posted by The Power
Where that last bit comes from ? I see where 5 comes from, but not
(2x - 3)^2

First, check that 5 - (2x - 3)^2 really equals what it must; second, in fantastic's post, just before he obtained this expression, he took out (1/4) from the parentheses but here he got;

[5/4 - (x - 3/2)^2] = [5/4 - ((2x - 3)/2)^2] = [5/4 - (1/4)(2x-3)^2] and now 1/4 goes out and etc.

Tonio
• Oct 7th 2009, 11:01 AM
The Power
Still dont see where 1/4 and 2x magically comes from
• Oct 7th 2009, 11:16 AM
tonio
Quote:

Originally Posted by The Power
Still dont see where 1/4 and 2x magically comes from

Ok, this already is high school algebra, so "get your hands dirty" and begin working on it (at the very least grab a pencil and a sheet of paper and write down both fantastic's message and mine to see clearly what's written there...and then begin doing maths!)

Tonio
• Oct 7th 2009, 01:16 PM
The Power
Ah, I see, he expanded out (x - 3/2) ^2 the multiplied, through -4 so techincally his simplifaction isn't needed just more easier.
• Oct 7th 2009, 01:26 PM
tonio
Quote:

Originally Posted by The Power
Ah, I see, he expanded out (x - 3/2) ^2 the multiplied, through -4 so techincally his simplifaction isn't needed just more easier.

I don't know what he did, but no need at all to expand anything:

(x - 3/2)^2 = ([2x-3]/2)^2 = (1/4)*(2x-3)^2

Tonio
• Oct 7th 2009, 03:17 PM
The Power
Do I use the trig. substitution of arcsin x ?
• Oct 7th 2009, 03:39 PM
tonio
Quote:

Originally Posted by The Power
Do I use the trig. substitution of arcsin x ?

I supsoe you can try substitution, but in fact:

(1/5)[1/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] =

(1/2Sqrt(5)))[(2/Sqrt(5))/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] ==>

the integral of the above equals (1/2Sqrt(5))*Arcsin[(2x-3)/sqrt(5)]

Tonio
• Oct 7th 2009, 04:07 PM
The Power
Mind throwing some math tags around those ?

Couldn't I use 2x - 3 = sin theta and solve for x, I tried that but its seems to complicate things
• Oct 7th 2009, 06:28 PM
mr fantastic
Quote:

Originally Posted by The Power
Mind throwing some math tags around those ?

Couldn't I use 2x - 3 = sin theta and solve for x, I tried that but its seems to complicate things

tonio has graciously walked you through basic algebra that a student attempting questions like the one you posted should have had no trouble with. So if you want math tags, I suggest you quote what tonio wrote and then insert the tags yourself into the quote. And the least you could do is to actually acknowledge the help you got by saying thankyou (or clicking on the Thanks button).

If you follow what's been posted in this thread then it should be clear that you have a standard form. I had hoped the basic substitution u = 2x - 3 would have made that obvious. I suggest you go back and review this material.
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