Question

Here's an interesting integral: the primitive of 1/sqrt( x(1-x) )

Answer 1

\[ \int \frac{dx}{\sqrt{x(1-x)}} \]


I have a solution for it, but would interested to see how others approach the problem. This is one case (among others!) where WolframAlpha is actually not very helpful.

Answer 2

I am not sure if what I have in mind is somehow useful for this kind of problem, but I was considering expending the term under the radical and then making a perfect square out of it. Afterwards a trig substitution.

Answer 3

Yep, completing the square would be the most apparent route.

Answer 4

ok ... so work it through

Answer 5

I am not too sure about it yet. Some of the signs bother me still

Answer 6

Yes, the signs are bizarre and things become problematic. I had to take things into complex numbers to make it easy to work through at first and then do some slightly non-standard (for an integral problem) manipulation to get back to a real function.

Answer 7

\[\int \frac{dx}{\sqrt{x^2-x}}\]
x² - x
x² - x + 1/4 - 1/4
(x - 1/2)² - 1/4
\[\int \frac{dx}{\sqrt{(x-\frac{1}{2})^2-\frac{1}{4}}}\]
Let x - 1/2 = 1/2 secu
dx = 1/2 secu tanu du
\[\int \dfrac{\frac{1}{2}\sec u\tan u}{\sqrt{(\frac{1}{2}\sec u)^2 - \frac{1}{4}}}du\\
\frac{1}{2}\int \dfrac{\sec u\tan u}{\sqrt{\frac{1}{4}\sec^2 u - \frac{1}{4}}}du\\
\int \dfrac{\sec u\tan u}{\sqrt{\sec^2 u - 1}}du\\
\int \dfrac{\sec u\tan u}{\sqrt{\tan^2 u}}du\\
\int \dfrac{\sec u\tan u}{\tan u}du\\
\int \sec u\;du\]
And from there, it's all a matter of subbing back.

Answer 8

For the Radical:

\[\Large \sqrt{-x^2+x-\frac{1}{4}+\frac{1}{4}} = \sqrt{-(x-\frac{1}{2})^2-\frac{1}{4}} \]

Answer 9

So I see where the signs get complicated.

Answer 10

Oops, looks like I made a mistake with the sign... Just a sec.

Answer 11

My Sith friend, it is \( \sqrt{x - x^2} \), which does make a difference.

Answer 12

in fact, it makes all the difference (-: ! I would go with a complex substitution now too. Let me see.

Answer 13

Here's the change to my original reply:
In the denominator, under the radical, you'd have
-x² + x + 1/4 - 1/4
-(x² - x + 1/4 - 1/4)
-((x - 1/2)² - 1/4)
1/2 - (x - 1/2)²

Let u = x - 1/2
du = dx

So the integral becomes
\[\int \dfrac{du}{\sqrt{\frac{1}{4}-u^2}}\]
Then, let u = 1/2 sint
du = 1/2 cost dt

I think this should fix that.

Answer 14

Unless I made some minor algebraic mistake again...

Answer 15

Looks like I did, but nothing really changes. In the fourth line, the 1/2 should be 1/4.

Answer 16

so your final answer comes out as (1/2).arcsin(2x-1), yes?

Answer 17

Nice.

Answer 18

I haven't worked it out completely, but that's what it looks like it'll turn out to be.

Answer 19

Yes, that's right. From there you can deduce this interesting result as well
\[ \int_0^1 \frac{dx}{\sqrt(x(1-x)} = \pi \]

***

More methods welcome.

Answer 20

I concur, Just that my substitution now (on a second tryI)

was different.

\[\Large \alpha = \sin^{-1}(2x-1) \]

Answer 21

\[ \int_0^1 \frac{dx}{\sqrt{x(1-x)}} = \pi \]

Answer 22

right, the (1/2) I had there was wrong.

Answer 23

Well, your title wasn't promising too much @JamesJ. A very interesting integral.

Answer 24

Here's my solution for the record. There's a slight bug towards the end with a minus sign. The final answer should be
\[ I = -2 \arctan\sqrt{\frac{1}{x} - 1} \]
http://openstudy.com/study#/updates/5112b975e4b0e554778a94c3

Answer 25

Which also suggests one of those truly arcane identities:
\[ \arcsin(2x - 1) + 2\arctan\sqrt{\frac{1}{x} - 1} = \frac{\pi}{2} \]

Answer 26

I evaluated the integral via substitution. It works great. :D\[\int\limits\limits \frac {dx}{\sqrt{x} \sqrt {1 - x}}, u = \sqrt {1-x}, -2du = \frac {dx}{\sqrt{1-x}}, \sqrt {x} = \sqrt {1 - u^2}\]\[\int\limits \frac {-2du}{\sqrt{1 - u^2}} = -2 \sin^{-1} (u) = -2 \sin^{-1} (\sqrt{1-x})\]

Answer 27

Yes, that's neat too. Good stuff.