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

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SUMMARY

The discussion focuses on evaluating the indefinite integral of sin(2x)/(1+(cos(2x))^2). The user initially attempts a substitution with u = 1 + (cos(2x))^2, leading to a derivative of du = -4(cos(2x))(sin(2x)) dx. However, an alternative substitution suggested by another user involves letting u = cos(2t), which connects to the derivative of arctan(u). This approach is deemed more effective for solving the integral.

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  • Understanding of trigonometric identities and functions
  • Familiarity with integration techniques, particularly substitution
  • Knowledge of derivatives, especially involving inverse trigonometric functions
  • Basic proficiency in calculus, specifically indefinite integrals
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  • Study the method of substitution in integration, focusing on trigonometric functions
  • Learn about the derivatives of inverse trigonometric functions, particularly arctan
  • Explore the properties of trigonometric identities and their applications in integration
  • Practice evaluating integrals involving products of trigonometric functions and polynomials
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Students studying calculus, particularly those focusing on integration techniques and trigonometric functions. This discussion is beneficial for anyone looking to enhance their understanding of evaluating complex integrals.

Jet1045
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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.
 
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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.
 
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?
 
Jet1045 said:
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) .
 

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