I am trying to understand a calculation made in a text book.
We have the following equation:
$A\frac{\mathrm{d^2}y}{\mathrm{d}x^2}=\mathrm{sin(y)}$
Then is it stated that multiplying both side with dy/dx and then integrating we must get:
$A \frac{1}{2} \big( \frac{\mathrm{d}y}{\mathrm{d}x} \big)^2=\mathrm{cos(y)}+C$.
I do not understand the calculation. Multiplying both sides with dy/dx , we get:
$A\frac{\mathrm{d^2}y}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{sin(y)}\frac{\mathrm{d}y}{\mathrm{d}x}$
Integrating with respect to dx on both sides we get:
$\int$ $A\frac{\mathrm{d^2}y}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x=\int \mathrm{sin(y)}\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$
which then can be written as
$\int$ $A\frac{\mathrm{d^2}y}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x=\int \mathrm{sin(y)}\mathrm{d}y$
Yielding:
$\int$ $A\frac{\mathrm{d^2}y}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x=cos(y)+C$.
I understand how he get the right hand side, but not the left hand side.
My guess is that maybe he is rewriting the left hand side as:
$\int$ $A\frac{\mathrm{d^2}y}{\mathrm{d}x^2} \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$=$\int$ $A\frac{\mathrm{d} }{\mathrm{d}x} \big( \frac{\mathrm{d}y}{\mathrm{d}x} \big) \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$=$\int$ $A\frac{\mathrm{d} }{\mathrm{d}x} \big( \frac{\mathrm{d}y}{\mathrm{d}x} \big)^2\mathrm{d}x$
I am not sure...