Integral of ((-6x)^2 + (9x^2 - 1)^2)^(1/2)

[plasmatic]

I was going some integral and came across this:

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut (-6x)^2\, +\, (9x^2\, -\, 1)^2\,}\, dx\)

Then the working putting it into http://www.integral-calculator.com/ as I was simplifying and factoring and coming out with the wrong answer I get:

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. . . . .\(\displaystyle \mbox{Problem:}\)

. . . . .\(\displaystyle \displaystyle \int\, \sqrt{\strut (9x^2\, -\, 1)^2\, +\, 36x^2\,}\, dx\)

. . . . . \(\displaystyle \mbox{Factor and simplify:}\)

. . . . .\(\displaystyle \displaystyle =\, \int\, (3x\, -\, i)(3x\, +\, i)\, dx\)

. . . . .\(\displaystyle \mbox{Expand:}\)

. . . . .\(\displaystyle \displaystyle =\, \int\, 9x^2\, +\, 1\, dx\)

. . . . .\(\displaystyle \mbox{Apply linearity:}\)

. . . . .\(\displaystyle \displaystyle =\, 9\, \int\, x^2\, dx\, +\, \int\, 1\, dx\)

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In the factoring stages. I haven't worked with complex numbers a lot, so I don't know where this comes from or if this is the only method to simplify this. I tried expanding normally and came to an integral that worked, but it did not give the right answer, so evidently wrong:

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 36t^2\, +\, (9t^2\, -\, 1)^2\, }\, dt\)

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 36t^2\, +\, 81t^4\, -\, 18t^2\, +\, 1\,}\, dt\)

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 81t^4\, +\, 18t^2\, +\, 1\,}\, dt\, =\, \)\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \left(t^2\, +\, \dfrac{18}{162}\right)^2\, dt\)

\(\displaystyle =\, \bigg[\, \dfrac{t^3}{3}\, +\, \dfrac{18}{162}t\,\bigg]_0^1\)

\(\displaystyle 81x^2\, +\, 18x\, +\, 1\)

\(\displaystyle \dfrac{-18\, \pm\, \sqrt{\strut 324\, -\, 324\,}}{162}\, =\, \dfrac{-18\, \pm\, \sqrt{\strut 0\,}}{162}\)

Can someone explain why and how this has been factored this way? And if there are any alternative methods to getting this answer?

 

[pka]

I was going some integral and came across this:

Can someone explain why and how this has been factored this way? And if there are any alternative methods to getting this answer?

I have no idea what whoever did that is trying. I would count it as correct, since this is basic algebra.
\(\displaystyle \large\sqrt{(-6x)^2+(9x^2-1)^2}=9x^2+1\)

 

[stapel]

I was going some integral and came across this:

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut (-6x)^2\, +\, (9x^2\, -\, 1)^2\,}\, dx\)

Then the working putting it into http://www.integral-calculator.com/ as I was simplifying and factoring and coming out with the wrong answer I get:

---

. . . . .\(\displaystyle \mbox{Problem:}\)

. . . . .\(\displaystyle \displaystyle \int\, \sqrt{\strut (9x^2\, -\, 1)^2\, +\, 36x^2\,}\, dx\)

. . . . . \(\displaystyle \mbox{Factor and simplify:}\)

. . . . .\(\displaystyle \displaystyle =\, \int\, (3x\, -\, i)(3x\, +\, i)\, dx\)

. . . . .\(\displaystyle \mbox{Expand:}\)

. . . . .\(\displaystyle \displaystyle =\, \int\, 9x^2\, +\, 1\, dx\)

. . . . .\(\displaystyle \mbox{Apply linearity:}\)

. . . . .\(\displaystyle \displaystyle =\, 9\, \int\, x^2\, dx\, +\, \int\, 1\, dx\)

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In the factoring stages. I haven't worked with complex numbers a lot, so I don't know where this comes from or if this is the only method to simplify this....

They skipped the step shown in the first reply, where they multiplied out and simplified to get 9x^2 + 1. This is a sum of squares, so it's not factorable in the real numbers. However, in the complexes, using "sqrt{-1} = i" (so i^2 = -1), we get:

. . . . .\(\displaystyle 9x^2\, +\, 1\, =\, (3x)^2\, -\, (-1)\, =\, (3x)^2\, -\, (i)^2\, =\, (3x\, +\, i)\, (3x\, -\, i)\)

I tried expanding normally and came to an integral that worked, but it did not give the right answer, so evidently wrong:

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 36t^2\, +\, (9t^2\, -\, 1)^2\, }\, dt\)

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 36t^2\, +\, 81t^4\, -\, 18t^2\, +\, 1\,}\, dt\)

\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \sqrt{\strut 81t^4\, +\, 18t^2\, +\, 1\,}\, dt\, =\, \)\(\displaystyle \displaystyle \int_0^1\, \)\(\displaystyle \left(t^2\, +\, \dfrac{18}{162}\right)^2\, dt\)

I don't understand your last step (nor the side computation, which looks like it might come from the Quadratic Formula, which is irrelevant to the factorization needed)...? Instead, try:

. . . . .\(\displaystyle 81t^4\, +\, 18t^2\, +\, 1\, =\, (9t^2)^2\, +\, 2(9t^2)(1)\, +\, (1)^2\, =\, (9t^2\, +\, 1)^2\)

Then take the square root of the square, and continue.

 

[plasmatic]

I used the quadratic formula because it gives me a solution to the equation, which I then use to factor. I did that method with both substituting p = t^2 and p = 9t^2 in my working below. Upon closer inspection, doing that gives me the right answer for the latter substitution and the former gives the same answer divided by 9.