Getting wrong answer trying to evaluate $\int \frac {\sin(2x)dx}{(1+\cos(2x))^2}$ I'm trying to evaluate $$\int \frac {\sin(2x)dx}{(1+\cos(2x))^2} = I$$
Here's what I've got:
$$t=1+\cos(2x)$$
$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$
$$I = \int \frac 1{t^2} \cdot \frac {-1}2 dt = -\frac 12 \int t^{-2}dt = -\frac 12 \frac {t^{-2+1}}{-2+1}+C$$
$$=-\frac 12 \frac {t^{-1}}{-1}+C = \frac 1{2t} +C = \frac 1{2(1+\cos(2x)}+C$$
I need to find what went wrong with my solution. I know the answer is correct but the line $dt=-2\sin(2x)dx$ concludes that $dx=\frac {dt}{-2}$ is false.  But why?  What should it be instead?

Perhaps simpler:
$$\begin{cases}1+\cos2x=1+2\cos^2x-1=2\cos^2x\\{}\\\sin2x=2\sin x\cos x\end{cases}\;\;\;\;\;\;\;\;\;\;\;\implies$$
$$\int\frac{\sin2x}{(1+\cos2x)^2}=\int\frac{2\sin x\cos x}{4\cos^4x}dx=-\frac12\int\frac{(\cos x)'dx}{\cos^3x}=-\frac12\frac{\cos^{-2}x}{-2}+C=$$
$$=\frac14\sec^2x+C\ldots\ldots\text{and your solution is correct, of course}$$
using that
$$\int\frac1{x^3}dx=-\frac12\frac1{x^2}+c\implies \int\frac{f'(x)}{f(x)^3}dx=-\frac1{2f(x)^2}+c$$
