Trigonometric Integral: Evaluating sin2x/(1+(cos2x)^2)

[Jet1045]

Homework Statement

The question asks me to evaluate the indefinite integral:

sin2x/(1+(cos2x)^2)

The (cos2x)^2 is read cos squared 2x, i just don't know how to put the squared before the 2x and not have it look confusing !

Homework Equations

The Attempt at a Solution

To be honest i have NOO clue where to start. Right now in class we are learning about inverse trig, hyperbolic trig, inverse hyperbolic trig, and the derivatives for each of them.
ANY hint on how to start this would be greatly appreciated.

 

[jbunniii]

A substitution seems like a reasonable way to start. You pretty much have two choices: either u = sin(2x) or u = cos(2x). Only one of these will take you very far.

 

[Jet1045]

How did i not see that...hhaha
THANKS!

OK!

SO, i set:
u = 1 + (cos2x)^2
du = -4(cos2x)(sin2x) dx
dx = du / -4(cos2x)(sin2x)

when i insert this though, i don't seem to get far.
is my derivative of 1+(cos2x)^2 correct?

 

[SammyS]

How did i not see that...hhaha
THANKS!

OK!

SO, i set:
u = 1 + (cos2x)^2
du = -4(cos2x)(sin2x) dx
dx = du / -4(cos2x)(sin2x)

when i insert this though, i don't seem to get far.
is my derivative of 1+(cos2x)^2 correct?

That's not the substitution suggested by jbunniii.

He suggested letting u = cos(2t) . This should give something that's related to the derivative of the arctan(u).

NOT u = 1+cos²(2t) .