Inverse integral manipulation

Question

Why is the following true:

If $\rho = B^{-1}(t)$ where

$$ B(t) = \int_0^t \frac{1}{g(\gamma(s))}ds $$

then

$$ t = \int_0^\rho \frac{1}{g(\gamma(s))}ds $$

I know it must be something fundamental, but I'm unable to figure out the sequence of operations.

Thanks.

Answer 1

It seems obvious. Assuming $B$ is invertible in suitable interval, then $B(t)=\int_0^t \frac{ds}{g(\gamma(s))}\Rightarrow\:B(B^{-1}(t))=\int_0^{B^{-1}(t)} \frac{ds}{g(\gamma(s))}\Rightarrow t=\int_0^{\rho(t)} \frac{ds}{g(\gamma(s))}$ .

By the way its better to use $\rho(t)$ instead of $\rho$.