how to reduce the integral into hypergeometric function? The equation is 
$\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi
^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},%
\frac{1}{2};2;\Pi ^{2})$,
where $\Pi \le 1$.
How to reduce the left integral into the right hypergeometric fucntion? Is there anyone can solve this problem? Thank you.
Thank you for @T. Amdeberhan. I learn general binomial expansion from your answer.
Your result is more general. After some modifications, the result is
\begin{eqnarray*}
&&\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi
^{2}-2\Pi \sqrt{1-y^{2}}}\right)  \\
&=&\frac{\pi }{2}\frac{\Pi }{\sqrt{1+\Pi ^{2}}}\ _{2}F_{1}\left( \frac{1}{4},%
\frac{3}{4};2;\left( \frac{2\Pi }{1+\Pi ^{2}}\right) ^{2}\right) ,
\end{eqnarray*}
which is valid for any value of $\Pi $. But for $\Pi \leq 1$, how to
transform the result into
$
\int_{0}^{1}dy\left( \sqrt{1+\Pi ^{2}+2\Pi \sqrt{1-y^{2}}}-\sqrt{1+\Pi
^{2}-2\Pi \sqrt{1-y^{2}}}\right) =\frac{\pi \Pi }{2}\ _{2}F_{1}(-\frac{1}{2},%
\frac{1}{2};2;\Pi ^{2})
$? 
It is seems that we should expand the integrad with respect to $\Pi $.
Should we use multinomial expansion? But how to deal with the double
summation? Could we prove the equality of the two hypergeometrical functions
directly for $\Pi \leq 1$?
 A: Letting $\beta=\frac{2\Pi}{1+\Pi^2}$, write the integrand as
$$\sqrt{1+\Pi^2}\left(\sqrt{1+\beta \sqrt{1-y^{2}}}-\sqrt{1-\beta \sqrt{1-y^{2}}}\right).\tag1$$
Use the binomial expansion to express (1), after combining two infinite series, in the form
$$2\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\beta^{2n+1}\int_0^1(1-y^2)^n\sqrt{1-y^2}dy.\tag2$$
Standard integral evaluations enable to compute the integral, hence (2) becomes
$$\begin{align}
&2\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\beta^{2n+1}\binom{2n+2}{n+1}\frac{\pi}{2^{2n+3}} \\
=& \frac{\beta\pi}4\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{1/2}{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}2\right)^{2n} \\
=& \frac{\beta\pi}4\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{4n}{2n}\frac1{(2n+1)2^{4n+1}}\binom{2n+2}{n+1}\left(\frac{\beta}2\right)^{2n} \\
=& \frac{\beta\pi}8\sqrt{1+\Pi^2}\sum_{n\geq0}\binom{4n}{2n}\frac1{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}8\right)^{2n} \\
=& \frac{\pi\Pi}{4\sqrt{1+\Pi^2}}\sum_{n\geq0}\binom{4n}{2n}\frac1{2n+1}\binom{2n+2}{n+1}\left(\frac{\beta}8\right)^{2n} \\
=& \frac{\pi\Pi}{4\sqrt{1+\Pi^2}}\sum_{n\geq0}\frac{(4n)!}{(2n)!(n+1)!n!}\left(\frac{\Pi}{4+4\Pi^2}\right)^{2n}.
\end{align}$$
