C4 integration - two "right" answers??

Okay, well actually there is only one but HEAR ME OUT plz.

So I was trying to solve

I did the idea of integrating a first order differential to get me back to the original function (+C!!!!!)

So I did y=-1/4 * csc^2(2x)

which when I put into WA, is identical to the integrand - But it's not the answer.

Why?!

Okay, well actually there is only one but HEAR ME OUT plz.

So I was trying to solve

I did the idea of integrating a first order differential to get me back to the original function (+C!!!!!)

So I did y=-1/4 * csc^2(2x)

which when I put into WA, is identical to the integrand - But it's not the answer.

Why?!

I haven't checked this, but I'll assume you have everything right so far. When you say this isn't the answer, do you mean it is not the given answer? As in the mark schme or text book answer? Because if the differential of your answer is the original integrand, then it is correct. I expect that you will need to do some trig identity work to verify that your answer and their answer are, in fact, the same.

Note cosec^2 and cot^2 only differ by a constant, and you get an arbitrary constant when you integrate...

I presume where you mention "WA" you mean Wolfram Alpha.

Wolfram Alpha gives the antiderivative as $-(1/4) \cot ^{2} (2x) + C,$ (where C is a constant).

As DFranklin says, cot^{2} (x) and cosec^{2} (x) only differ by a constant.

Note, as under "Alternate forms of the integral", it lists:

$- \dfrac{ \cos^{2} (2x) }{ 4 \sin^{2} (2x) } + C.$

As I mentioned, yeah.

But the point still remains that if you differentitate csc^2 (2x) * -1/4 you get the integrand.

So maybe that +C at the end takes care of it?