Questions about reduction of elliptic integrals and general calculus

Question

I am working on a numerical problem and trying to simplify elliptic integrals to reduce computational costs.
I found a previous question that is similar to my case, but I don't understand part of it.
I'll take that part out and talk about it. $$H_n(k)=\int_0^1 \frac{t^{2n}dt}{\sqrt{1-t^2}\sqrt{1-kt^2}}$$ Introducing for simplicity function $\rho_k=\sqrt{1−kt^2}$, $$H_n(k)=\int_0^1 \frac{t^{2n}}{\rho_1 \rho_k} dt=-\int_0^1 \frac{t^{2n}}{\rho_k}d\rho_1=\int_0^1 \rho_1d \frac{t^{2n}}{\rho_k}...$$ I don't understand the equation transformation for the third equal sign in the above equation.
I think there is a deformation of $\int_0^1 f(x)dx=-\int_0^1xdf(x)$, but I don't understand why it is valid.
I would appreciate it if you could tell me.

Answer 1

$$H_n(k)=\int_0^1 \frac{t^{2n}}{\rho_1 \rho_k} dt=-\int_0^1 \frac{t^{\color{red}{2n-1}}}{\rho_k}d\rho_1=\int_0^1 \rho_1d \frac{\color{red}{t^{2n-1}}}{\rho_k}...$$ The first equality: $$\dfrac {d \rho_1}{dt}=\dfrac {d \sqrt {1-t^2}}{dt}$$ $$\dfrac {d \rho_1}{dt}=-\dfrac t {\sqrt {1-t^2}}$$ $$\dfrac {d \rho_1}{dt}=-\dfrac t {\rho_1}$$ $$\implies {d \rho_1}=-\dfrac t {\rho_1}dt$$ The second equality is just an integration by part: $$-\int_0^1 \frac{t^{\color{red}{2n-1}}}{\rho_k}d\rho_1=\Bigg| - \frac{t^{{2n-1}}}{\rho_k}\rho_1\Bigg|_0^1+\int_0^1 \rho_1 \left ( \frac{t^{{2n-1}}}{\rho_k}\right)'dt$$ $$-\int_0^1 \frac{t^{{2n-1}}}{\rho_k}d\rho_1=\int_0^1 \rho_1 d\left ( \frac{t^{{2n-1}}}{\rho_k}\right)$$