Question

Find the indefinite integral of 1/(x^(1/2)(1-x)^(1/2))dx

Answer 1

I was thinking of using (1-x)^(1/2)

Answer 2

This is a bizarre integral. Here's what I've got:

\[ \frac{1}{\sqrt{x(1-x)}} = -i\frac{1}{\sqrt{x(x-1)}} = -2i \left( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x-1}}\right) \frac{1}{\sqrt{x} + \sqrt{x-1}}\]
and hence integrating
\[ I \ (for \ integral) \ = -2i.\ln(\sqrt{x} + \sqrt{x-1}) + C\]
which certainly looks bizarro. But as the integrand only makes sense as a real number for
\( x \in (0,1) \), we can rewrite this integral as
\[ I = -2i.\ln(\sqrt{x} + i\sqrt{1-x}) + C\]

Now this complex number being evaluated by the ln function, call it \( z = \sqrt{x} + i\sqrt{1-x} \) clearly has modulus 1 and its argument is \( \arctan\sqrt{\frac{1}{x} - 1} \).

Hence \( \ln(z) = \ln|z| + i\arg(z) = \ln 1 + i \arctan\sqrt{\frac{1}{x} - 1} = i \arctan\sqrt{\frac{1}{x} - 1} \)
and therefore
\[ I = 2\arctan\sqrt{\frac{1}{x} - 1} + C \]

Using this answer, we can see that
\[ \int_0^1 \frac{1}{\sqrt{x(1-x)}} dx = \pi \]

Knowing now this final real function, I"m sure there is a way to derive it not using complex numbers.

***

Nice problem!

Answer 3

Dropped a minus sign above for I

Answer 4

@torr19, for more discussion and another approach, see
http://openstudy.com/study#/updates/5112c4e3e4b0e554778a9872

Answer 5

try the substitution x= u^2, dx = 2u du
\[ \int \frac{1}{\sqrt{x(1-x)}} = \int \frac{2 u du }{u \sqrt{1-u^2}}=2 \int \frac{du }{\sqrt{1-u^2}}\]

that is a standard integral, but if we insist on doing it with substitution


the integral is
\[ 2 \int \frac{cos Z}{cos Z} dZ = 2\int dZ = 2Z \]
substituting back in we get
\[ 2 \sin^{-1} u = 2 \sin^{-1} \sqrt{x}\]