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 $0

Question

I 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.

Answer 1

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.