Since $$\pi_{(2)}(x)=\sum_{i=1}^{\pi(x^{1/2})}\left(\pi\left(\dfrac{x}{\text{p}_i}\right)-i+1\right),$$ where $\pi_{(2)}(x)$ denotes the semiprimes and $\text{P}_i$ is the $i$th prime, an asymptotic of the semiprimes may be given by $$\pi_{(2)}(x)\sim R\left(\dfrac{x}{2}\right)+\sum_{i=2}^{R(x^{1/2})}\left(R\left(\dfrac{x}{R^{(-1)}(i)}\right)-i+1\right),$$ where$$\operatorname{li}(x)=\int_{0}^{\infty}\dfrac{1}{\log(t)}\operatorname{d}t,$$ $$R(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\operatorname{li}(x^{1/n}),$$ and the inverse of $R(x)$ is approximated by the Halley iterations $$p(n,i)=n-\dfrac{2n\log(n)(R(n)-i)}{2n+R(n)-i} $$ $$\Rightarrow R^{(-1)}(i)\approx p^{(3)}(i\log(i),i)\sim\text{P}_i.$$ This is a step function however, which can be smoothed in a piecewise fashion with a sawtooth function at $R(x^{1/2})=\text{Int.}$ as follows:
Is this smoothing valid for all $x$?
