I have a question:
Right before this question, in a previous part, they had me find the derivative of $\frac{1}{\sin^2x}$, which I found to be $\frac{-2\cos\theta}{\sin^3\theta}$.
My problem is with the integration part (I've separated my variables). I tried u-substitution with $u = \sin^4\theta$, and also integration by parts, but neither got anywhere. I seem to have gotten into an loop with integration by parts, and u-substitution ended when I still had a $\cos\theta$ in my numerator after all substitutions/cancelling etc. So now, having spent a considerable amount of time on this problem, I don't see how to integrate the right hand side after separating variables. The right hand side is:
$$\left(\frac{-1}{\sin^2\theta}\right)\left(\frac{\cos \theta}{\sin \theta}\right) d\theta$$
Best Answer
As @player3236 notes, in the problem's original notation $xdx=-\frac{\cos\theta}{\sin^3\theta}d\theta$. With $s=\sin\theta$,$$x^2=\int 2xdx=-\int 2s^{-3}ds=s^{-2}+C=\csc^2\theta+C.$$I'll leave you to find $C$.