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I'm a little rusty with my integrals, how may I evaluate the following:

$$ \int_0^1{\frac{y}{\sqrt{y(1-y)}}dy} $$

I've tried:

$$ \int_0^1{\frac{y}{\sqrt{y(1-y)}}dy} = \int_0^1{\sqrt{\frac{y}{1-y}} dy} $$

Make the substitution z = 1-y

$$ = \int_0^1{\sqrt{\frac{1-z}{z}} dz} = \int_0^1{\sqrt{\frac{1}{z} - 1} dz} $$

Which seems simpler but I have not managed to evaluate it. All help appreciated (just a few pointers would be appreciated just as much as the full solution).

Thank you

EDIT:

Solution following Ng Hong Wai suggestion:

At $$ \int_0^1{\sqrt{\frac{y}{1-y}} dy} $$

We make the substitution $y = sin^2(\theta)$, so $dy = 2sin(\theta)cos(\theta)d\theta$. This results in:

$$ \int_0^\frac{\pi}{2}{2 sin^2(\theta)d\theta} = \int_0^\frac{\pi}{2}{[1-cos(2\theta)]d\theta} = [x-\frac{1}{2}sin(2\theta)]_0^\frac{\pi}{2} = \frac{\pi}{2} $$

Solution according to M. Strochyk's suggestion:

$$ \int_0^1{\sqrt{\frac{y}{1-y}} dy} = 2 \int_0^\infty{\frac{u^2}{(1+u^2)^2}du} $$

By decomposing into partial fractions this becomes:

$$ 2 \int_0^\infty{\frac{1}{1+u^2}du} - 2 \int_0^\infty{\frac{1}{(1+u^2)^2}du} $$

The first term integrates directly to $arctan(u)$ while the second can be evaluated by making the substitution $u=tan(x)$ and then applying the double-angle formula for cosine. The result is (unsurprisingly):

$$ 2[arctan[u]]_0^\infty -[x + \frac{1}{2}sin(2x)]_0^\frac{\pi}{2} = \frac{\pi}{2} $$

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  • $\begingroup$ Complete the square? $\endgroup$ – Jyrki Lahtonen Jan 7 '13 at 15:41
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At $$\int_{0}^{1}{\sqrt{\frac{y}{1-y}}}dy,$$ you can use substitution $y=sin^2\theta$

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  • $\begingroup$ Am I correct in saying that I could not use this substitution if my limits of integration went outside of [0, 1]? $\endgroup$ – Amos Joshua Jan 9 '13 at 9:54
  • $\begingroup$ @AmosJoshua : yes $\endgroup$ – Idonknow Jan 9 '13 at 9:57
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By using beta function $$\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}=B(3/2,1/2)=\frac{\Gamma(3/2)\Gamma(1/2)}{\Gamma(3/2+1/2)}=\frac{\Gamma^2(1/2)}{2}=\frac{\pi}{2}$$

Q.E.D.

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Substitute $u=\sqrt{\dfrac{y}{1-y}}.$ Then $$\dfrac{y}{1-y}=\dfrac{y-1+1}{1-y}=-1+\dfrac{1}{1-y}=u^2, \\ ( {0}< {y} <{1} \Leftrightarrow {0}< {u}<{+\infty}),\\ \dfrac{1}{1-y}=1+u^2, \\ 1-y=\dfrac{1}{1+u^2}, \\ y=1-\dfrac{1}{1+u^2}, \\ dy=\dfrac{2u}{(1+u^2)^2}\, du, $$ so $$\int\limits_0^1{\sqrt{\dfrac{y}{1-y}} dy}=\int\limits_0^{+\infty}{\dfrac{2u^2}{(1+u^2)^2}}\, du=2\int\limits_0^{+\infty}{\dfrac{u^2}{(1+u^2)^2}}\, du$$ Last integral can be calculated by decomposition into partial fractions.

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  • $\begingroup$ A little more work but thank you for this suggestion as well, I was just as rusty with partial fractions $\endgroup$ – Amos Joshua Jan 9 '13 at 10:14

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