This is a differential equation of SHM from my book. $$\frac{d^2x}{dt^2}=-\omega^2\times x$$ Both sides is multiplied by $\displaystyle 2\frac{dx}{dt}$ for simplification. And now, $$ 2\times \frac{dx}{dt} \frac{d^2x}{dt^2}=-2\omega^2\times x \frac{dx}{dt}$$
Book integrates both sides to simplify it further out of thin air.
I did managed to integrate the RHS but i have no idea how the integration in LHS works. Can someone point it out plz.
I don't know how to integrate it. can someone explain me how I can get LHS: $$\Bigl(\frac{dx}{dt}\Bigr)^2 = -\omega^2 + C$$
It might be stupidly simple but I couldn't find it online somehow.
Expanding on the comment by @LutzLehmann, the chain rule says that for some function $v$ of $t$, $$ \frac{d}{dt} \bigl(v^2\bigr) = 2v\frac{dv}{dt}, $$ which becomes $$ \frac{d}{dt} \biggl(\Bigl(\frac{dx}{dt}\Bigr)^2\biggr) = 2\frac{dx}{dt} \frac{d^2x}{dt^2}, $$ after the substitution $\displaystyle v = \frac{dx}{dt}$.
But how do we reverse-engineer the chain rule when faced with an integration problem? In other words, how would we integrate this expression without recognizing the integrand as a derivative a priori? This is essence of substitution.
With the substitution $\displaystyle v = \frac{dx}{dt}$, we have $$ dv = \frac{d}{dt} \biggl(\frac{dx}{dt}\biggr) \, dt = \frac{d^2x}{dt^2}\frac{dx}{dt} \, dt = \frac{d^2x}{dt^2} \, dx $$ so $$ \int 2\,\frac{dx}{dt} \frac{d^2x}{dt^2} \, dx = \int 2v \, dv = v^2 + C = \Bigl(\frac{dx}{dt}\Bigr)^2 + C $$