Help to find skipped step in integration

[johnpjust]

In a textbook I'm reading, a step in an example problem was skipped during integration, and I'm just not quite seeing how to get from "point A to B" in this.

(1) d(R)/(2*cos(x)) = (R*d(x))/(sin(x))

(2) d(R)/R = 2*d(sin(x))/(sin(x))

*Integrate* (here is where I get lost -- what is happening here?)

(3) R = (sin(x))^2

 

[arildno]

Well, you have:
$$\frac{dR}{R}=2\frac{d(\sin(x))}{\sin(x}}$$
Integrating this yields:
$$\ln|R|-\ln|R_{0}|=2*(\ln|\sin(x)|-\ln|\sin_{x}_{0}|)$$
Now, assuming that $|R_{0}|=|\sin(x_{0})|=1$*, we get:
$$\ln|R|=2\ln|\sin(x)|$$

I'm sure you manage the last step on your own, in the event that R>0

*In the general case, we get:
$$R=K\sin^{2}(x}$$
where K is some arbitrary constant.

 

[johnpjust]

Thanks...apparently I couldn't think straight last night because that is really obvious now haha.

I didn't realize though that the constant 'K' in the general case will be there. I knew from the context of the problem that there would have to be a constant there, but I guess I didn't know how that got there either (the solution actually has 'K' in it in the book).

How does - ln (|sin(x_0)|^2 + |R_0|) translate to the constant 'k' in the general case? Is it a property of logrithms?

 

[AlephZero]

How does - ln (|sin(x_0)|^2 + |R_0|) translate to the constant 'k' in the general case? Is it a property of logrithms?

$$\ln(R) - \ln(R_0) = 2 (\ln(\sin x) - \ln (\sin x_0))$$

$$\ln(R/R_0)) = 2 \ln(\sin x/\sin x_0)$$

$$R/R_0 = \sin^2 x / \sin^2 x_0$$

$$R = (R_0 / \sin^2 x_0) \sin^2 x$$