An identity of an Elliptical Integral 
Suppose $0<k<1$ and $\displaystyle
 K(k)=\int_0^1\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-k^2x^2)}}$. Let
   $\tilde{k}$ be $\tilde{k}^2=1-k^2$. Show that $$\displaystyle
 K(k)=\frac{2}{1+\tilde{k}}K\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)$$

There's a hint in Stein's Complex Analysis which is this change of variable : $x=\dfrac{2t}{1+\tilde{k}+(1-\tilde{k})t^2}$.
 A: For me, the only way I can remember this sort of complicated identities is through the relationship between the 
complete elliptic integral of the first kind
 $K(k)$ and the corresponding arithmetic-geometric mean.
$$K(k) = \int_0^\frac{\pi}{2} \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}} = \frac{\pi}{2\text{AGM}( 1, \sqrt{1-k^2})}\tag{*1}$$
Consider following integral
$$I(a,b) = \int_0^{\frac{\pi}{2}}
\frac{d\theta}{\sqrt{a^2\cos^2\theta + b^2\sin^2\theta}}$$
Introduce $x = b\tan\theta$, we can rewrite it as
$$I(a,b) = \int_0^{\frac{\pi}{2}} \frac{d\tan\theta}{
\sqrt{(1+\tan^2\theta)(a^2+b^2\tan^2\theta})}
= \int_0^\infty \frac{dx}{\sqrt{(x^2 + a^2)(x^2+b^2)}}\tag{*2}$$
Substitute $x$ by $\sqrt{ab} t$, we have
$$I(a,b) 
= \frac{1}{\sqrt{ab}}\int_0^\infty \frac{dt}{\sqrt{t^4 + \left(\frac{a}{b}+\frac{b}{a}\right)t^2 + 1}}
= \frac{1}{\sqrt{ab}}\int_0^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t}
$$
Notice the last integrand is invariant under transform $\displaystyle\;t \leftrightarrow \frac{1}{t}$. If we introduce two more variables $s$ and $y$ such that
$$s = \frac12 (t - t^{-1}) = \frac{y}{\sqrt{ab}}$$ and using the fact $$\frac{dt}{t} = \frac{d(t - t^{-1})}{t + t^{-1}} = \frac{ds}{\sqrt{s^2+1}}$$
We can rewrite $I(a,b)$ as
$$
\frac{2}{\sqrt{ab}}\int_1^\infty \frac{1}{\sqrt{ ( t - t^{-1})^2 + \frac{(a+b)^2}{ab}}}\frac{dt}{t}
= \frac{2}{\sqrt{ab}}\int_0^\infty 
\frac{ds}{\sqrt{\left(4s^2 + \frac{(a+b)^2}{ab}\right)(s^2+1)}}\\
= \int_0^\infty 
\frac{dy}{\sqrt{\left(y^2 + \left(\frac{a+b}{2}\right)^2\right)(y^2 + ab)}}
$$
Compare this with $(*2)$, we obtain an important identity:
$$I(a,b) = I\left(\frac{a+b}{2}, \sqrt{ab}\right)$$
This means $I(a,b)$ is invariant if we replace $(a,b)$ by their AM and GM.
Start with any pair of numbers $a,b$, it is well known if you repeat taking AM/GM of them,
the pairs will ultimately converge to a single number. This is called the arithmetic geometric mean of $a$ and $b$ and usually denoted as  $\text{AGM}(a,b)$. If one replace $a$, $b$ by this AGM in the definition of $I(a,b)$, we obtain
$$I(a,b) = \frac{\pi}{2\text{AGM}(a,b)}$$
Together with the obvious identity $K(k) = I(1,\sqrt{1-k^2})$, we immediately obtain $(*1)$.
Using these tools and notice $\text{AGM}(a,b)$ is homogenous. i.e. 
$$\text{AGM}(\lambda a, \lambda b) = \lambda \text{AGM}(a,b) \quad\implies\quad I(\lambda a, \lambda b) = \frac{1}{\lambda} I(a,b),$$ 
the desired identity follows immediately.
$$
K(k) 
= I(1,\tilde{k}) = I\left(\frac{1+\tilde{k}}{2},\sqrt{\tilde{k}}\right) = \frac{2}{1+\tilde{k}} I\left(1,2\frac{\sqrt{\tilde{k}}}{1+\tilde{k}}\right)\\
= \frac{2}{1+\tilde{k}} K\left( \sqrt{1 - \frac{4\tilde{k}}{(1+\tilde{k})^2}} \right)
= \frac{2}{1+\tilde{k}} K\left( \frac{1-\tilde{k}}{1+\tilde{k}} \right)
$$
A: $(*)\displaystyle\frac{\mathrm{d}x}{\mathrm{d}t}=2\times\frac{1+\tilde{k}+(1-\tilde{k})t^2-2(1-\tilde{k})t^2}{[1+\tilde{k}+(1-\tilde{k})t^2]^2}=2\frac{1+\tilde{k}-(1-\tilde{k})t^2}{[1+\tilde{k}+(1-\tilde{k})t^2]^2}$.
$(*)\displaystyle\sqrt{1-x^2}=\frac{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4+2k^2t^2-4t^2}}{1+\tilde{k}+(1-\tilde{k})t^2}=\frac{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2\tilde{k}^2t^2-2t^2}}{1+\tilde{k}+(1-\tilde{k})t^2}$.
$(*)\displaystyle\sqrt{1-k^2x^2}=\frac{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4+2k^2t^2-4k^2t^2}}{1+\tilde{k}+(1-\tilde{k})t^2}=\frac{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2k^2t^2}}{1+\tilde{k}+(1-\tilde{k})t^2}$.
Now by reparameterizing the Integral :
$\begin{align*}
\displaystyle K(k)&=2\int_0^1\frac{1+\tilde{k}+(1-\tilde{k})t^2}{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2\tilde{k}^2t^2-2t^2}}.
\frac{1+\tilde{k}+(1-\tilde{k})t^2}{\sqrt{(1+\tilde{k})^2+(1-\tilde{k})^2t^4-2k^2t^2}}\\
&\qquad.\frac{1+\tilde{k}-(1-\tilde{k})t^2}{[1+\tilde{k}+(1-\tilde{k})t^2]^2}.\mathrm{d}t\\
&=\frac{2}{1+\tilde{k}}\int_0^1\frac{1}{\sqrt{1+(\frac{1-\tilde{k}}{1+\tilde{k}})^2t^4-\frac{2\tilde{k}^2t^2+2t^2}{\color{red}{(1+\tilde{k})^2}} }}.
\frac{1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2}{\sqrt{1+(\frac{1-\tilde{k}}{1+\tilde{k}})^2t^4+\frac{2\tilde{k}^2t^2-2t^2}{\color{red}{(1+\tilde{k})^2}}}}\mathrm{d}t
\end{align*}$
Here is the Complete proof after achille hui's correction:
$\begin{align*}
\displaystyle LHS&=\frac{2}{1+\tilde{k}}\int_0^1 \frac{1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2}
{\sqrt{(1-t^2)\left(1-\frac{1-\tilde{k}}{1-\tilde{k}}t^2\right)}.\sqrt{\left(1-(\frac{1-\tilde{k}}{1+\tilde{k}})t^2\right)^2}}\mathrm{d}t\\
&=\displaystyle\frac{2}{1+\tilde{k}}\int_0^1 \frac{\mathrm{d}t}
{\sqrt{(1-t^2)\left(1-\frac{1-\tilde{k}}{1-\tilde{k}}t^2\right)}}\\
&=\displaystyle\frac{2}{1+\tilde{k}}K\left(\frac{1-\tilde{k}}{1+\tilde{k}}\right)\tag{QED}
\end{align*}$
