Let's assume that $f:[1,2]\to\mathbb{R}$ and $g:[1,2]\to\mathbb{R}$ are both Riemann-integrable functions. Then we know that multiplying both functions,i.e. $fg= f(x)g(x)$, is also a Riemann-integrable function. However, it is not true that $$\int_1^2f(x)g(x)dx = \int_1^2f(x)dx \int_1^2g(x)dx$$
We can take for example $1=x\frac{1}{x}$ and see: $$1=\int_1^21dx = \int_1^2 x\frac{1}{x}dx\neq\int_1^2xdx \int_1^2\frac{1}{x} dx =1.5 \ln(2)$$
If I try to calculate the Riemann-integral via Darboux sums and take an arbitrary partition $P$ of $[1,2]$, then on each subinterval $[t_{i-1},t_i]$ of $P$ I get:
$$ 1=\inf\limits_{x\in[t_{i-1},t_i]}(f)\inf\limits_{x\in[t_{i-1},t_i]}(g)\leq \inf\limits_{x\in[t_{i-1},t_i]}(fg)\leq \sup\limits_{x\in[t_{i-1},t_i]}(f)\sup\limits_{x\in[t_{i-1},t_i]}(g)=1, $$ because $f$ is the inverse of $g$.
Refining the partition yields the integral and would lead to the false conclusion that $$ 1=\int_1^2 x\frac{1}{x}dx=\int_1^2xdx \int_1^2\frac{1}{x} dx=1 $$ Where is my mistake?
Your problem, as some users have already pointed out, is that you know that $1=\inf_{x\in[t_{i-1},t_i]}(fg)$ (since $fg$ is constantly 1), but not (as you claim) that $1$ is equal to the product of the infima.