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?

Your only error was where you said
$$dt=-2\sin(2x)dx \implies dx = \frac {dt}{-2}$$
You should have said
$$dt=-2\sin(2x)dx \implies \sin(2x)dx = \frac {dt}{-2}$$
You need the $\sin(2x)$ because the original integral has it in the numerator:
$${\sin(2x)dx\over(1+\cos(2x))^2}={1\over(1+\cos(2x))^2}(\sin(2x)dx)={1\over t^2}\cdot{dt\over-2}$$
Everything else you did was fine, if perhaps a little overly elaborated.
