Prove $\int_0^{\pi/2}\frac{1}{\sqrt{(1+t)^2-4t\sin^2{u}}}du \equiv \int_0^{\pi/2}\frac{1}{\sqrt{1-t^2\sin^2{u}}}du$ for $0I am trying to prove that
$$\int_0^{\pi/2}\frac{1}{\sqrt{(1+t)^2-4t\sin^2{u}}}du \equiv \int_0^{\pi/2}\frac{1}{\sqrt{1-t^2\sin^2{u}}}du\text{ for }0<t<1.$$
Context: In another question, I am trying to make "Attempt $2$" work (after letting $x=2u)$. That is, I want to show that LHS here is increasing in $t$ for $0<t<1$. When playing with graphs, I accidentally discovered that LHS seems to be identical to RHS, which I know is increasing. But I do not know how to prove that LHS = RHS. Anyway, I think this question is interesting by itself.
Attempt: When $t=0$, we have LHS = RHS. Then I tried to show that $\frac{d}{dt}\text{LHS}=\frac{d}{dt}\text{RHS}$, but that seems to be even harder than the OP.
 A: Let $\displaystyle  K(t) := \int_0^{\pi/2} \frac{1}{\sqrt{1-t^2 \sin^2 u}} du$.
This is known as the Complete elliptic integral of the first kind.
Let $\displaystyle s := \sqrt{\frac{4t}{(1+t)^2}}$.
Then,
$$\int_0^{\pi/2} \frac{1}{\sqrt{(1+t)^2 -4t \sin^2 u}} du = \frac{1+\sqrt{1-s^2}}{2} K(s)$$
and
$$K(t) = K\left( \frac{2-s^2 + 2\sqrt{1-s^2}}{s^2}\right).$$
Hence it suffices to show that
$$\frac{1+\sqrt{1-s^2}}{2} K(s)  = K\left( \frac{2-s^2 + 2\sqrt{1-s^2}}{s^2}\right), \ 0 < s < 1. $$
This follows from a consideration of the characterization of $K$ by the arithmetic-geometric mean. See the Wikipedia page https://en.wikipedia.org/wiki/Elliptic_integral.
A: You can also obtain Landen's transformation from the Binomial theorem and Parseval's theorem.
$$\sum_{n=0}^{\infty}\frac{(2n)!k^n}{2^{2n}n!^2}=\frac{1}{\sqrt{1-k}}.$$
Parseval's:
$$\sum_{n=0}^{\infty}\left(\frac{(2n)!k^n}{2^{2n}n!^2}\right)^2=\frac{1}{\pi}\int_{-\pi}^{\pi}|\Re({\frac{1}{\sqrt{1-ke^{ix}}}})|^2dx,$$
since $\cos{(2x)}=1-2\sin^2{(x)}$:
$$\frac{1}{\pi}\int_{-\pi}^{\pi}|\Re({\frac{1}{\sqrt{1-ke^{ix}}}})|^2dx=\frac{1}{\pi}\int_{0}^{\pi}\frac{1}{\sqrt{-2k\cos{x}+k^2+1}}dx=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt{-4k\sin^2{(x)}+(k+1)^2}}dx,$$
since:
$$\sum_{n=0}^{\infty}\left(\frac{(2n)!k^n}{2^{2n}n!^2}\right)^2=\frac{2}{\pi}K(k) \hspace{.4cm}\implies K(k)=\int_{0}^{\pi/2}\frac{1}{\sqrt{-4k\sin^2{(x)}+(k+1)^2}}dx.$$
(Note that replacing $k$ with $-k$ leads to the same identity).
With the same approach $ke^{2ix}$ you can obtain:
$$(2) \hspace{.5cm}K(k)=\int_{0}^{\pi/2}\frac{1}{\sqrt{16k\sin^2{(x)}\cos^2{(x)}+(k-1)^2}}dx.$$
With $ke^{3ix}$:
$$(3) \hspace{.5cm}K(k)=\int_{0}^{\pi/2}\frac{1}{\sqrt{4k\sin^2{(3x)}+(k-1)^2}}dx,$$
with $0<k<1$. And so on. You can obtain a lot of transformations in a simple manner eluding amounts of trigonometric calculus.
