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 solution is good. The substitution $t=1+\cos2x$ gives
$$
dt=-2\sin2x\,dx
$$
so the integral becomes
$$
\int\frac{\sin2x}{(1+\cos2x)^2}\,dx=\int-\frac{1}{2}\frac{1}{t^2}\,dt=\frac{1}{2t}+C=\frac{1}{2(1+\cos2x)}+C
$$
Differentiating shows the correctness. However, there's no “canonical” answer; if you notice that $1+\cos2x=2\cos^2x$ you can as well rewrite the answer as
$$
\frac{1}{4\cos^2x}+C
$$
but also as
$$
\frac{1+\tan^2x}{4}+C
$$
or infinitely many variations thereof.
