Finding $ \int \sin^2(2x)/[1+\cos(2x)]dx$. I am surprisingly having a bit of difficulty with an indefinite integral which is interesting since the integral I solved before is
$$ \int \frac{1+\cos2x}{\sin^2(2x)} dx$$
The integral I am currently working on is 
$$ \int \frac{\sin^2(2x)}{1+\cos2x} dx$$
I first divided out giving: (this is the mistake I made early on, in this case you use the pythagorean identity for sin, and then cancel out $1+\cos2x$)
$$ \int \sin^2(2x)+ \sin^2(2x)\cdot \sec(2x) \,dx$$
Then I factored out the $\sin^2(2x)$ resulting in:
$$ \int \sin^2(2x)(1+ \sec(2x)) \,dx$$
Substituting for $u=2x$ and splitting the integral into two parts:
$$\frac{1}{2} \int \sin^2(u) du + \frac{1}{2} \int \sin^2(u)\sec(u) \,du$$ lets call this eq. 1.
Now, this is where I am having difficulty as 1.) dealing with even powers of sin and 2.) the $\sec(u)$ term is proving to be troublesome.
Another form of the above equation is:
$$ \frac{1}{2} \int \sin^2(u) \,du + \frac{1}{2} \int \sin(u)\tan(u) \,du$$ 
Some approaches I have tried are using different trigonometric identities e.g.
$$\sin^2(u) = \frac{1}{2} (1-\cos(2u))$$
however, this results in for eq. 1
$$ \frac{1}{4} \int 1-\cos(2u) \,du + \frac{1}{4} \int \frac{(1-\cos(2u)}{\cos(u)} du $$
Then I would have to use the cosine angle addition formula which quickly gets out of hand.
I understand there are different approaches to solving different indefinite integrals. The purpose of this problem is to only use substitutions.
Questions that I have are as follows, 1.) is it possible to continue along with the steps I have taken?, 2.) or must I do an entirely different substitution at the beginning. Sorry for the long post and thank you for your time. 

Note: While it is true that $\dfrac {a+b}c = \dfrac ac + \dfrac bc$,
$$\dfrac a{b+c}\neq \dfrac ab + \dfrac ac$$
In particular 
$$ \int \frac{\sin^2(2x)}{1+\cos2x} dx \neq \int \sin^2(2x)+ \sin^2(2x)\cdot \sec(2x) dx$$
