How Does U-Substitution Using Tan(x/2) Simplify Trigonometric Integrals?

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Homework Help Overview

The discussion revolves around the use of the substitution \( u = \tan(x/2) \) to simplify trigonometric integrals, specifically focusing on deriving expressions for \( \sin x \) and \( \cos x \) in terms of \( u \). The original poster is attempting to understand how to express \( \sin x \) and \( \cos x \) using this substitution to facilitate solving the integral \( \frac{\sin x}{\sin x + \cos x} \).

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivation of \( \sin x = \frac{2u}{u^2 + 1} \) and \( \cos x = \frac{1 - u^2}{u^2 + 1} \) from the substitution \( u = \tan(x/2) \). Questions arise regarding the steps needed to achieve these expressions and the reasoning behind the presence of \( u^2 + 1 \) in the denominators. Some participants suggest using trigonometric identities and drawing triangles to visualize the relationships.

Discussion Status

The discussion is ongoing, with some participants expressing confusion about the derivation process while others provide insights and alternative approaches. There is a mix of understanding, with at least one participant indicating they have grasped the concept after receiving clarification.

Contextual Notes

Participants are working within the constraints of a homework assignment that requires the use of specific trigonometric identities and substitutions. There is an emphasis on understanding the underlying concepts rather than simply applying formulas.

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Homework Statement



Im trying to figure out how to get sinx=2u/(u^2+1) and cosx=(1-u^2)/(u^2+1). In order to solve the problem integral sinx/(sinx+cosx)

Homework Equations


use tan(x/2) to solve the problem.

The Attempt at a Solution


all I can get to is u=sin(x/2)/cos(x/2)-------2u=sinx/cosx
I don't understand how we can get the u^2+1 on the bottom. All we need is the give the u-substitution but i donno how to do it.
 
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So, we let [itex]u=\tan(x/2)[/itex]. Then we need to evaluate

[tex]\frac{2u}{1+u^2}=\frac{2\tan(x/2)}{1+\tan^2(x/2)}[/tex]

Try to prove the following formula first:

[tex]1+\tan^2(\alpha)=\frac{1}{\cos^2(\alpha)}[/tex]
 
I just want to know how you get sinx=2u/(u^2+1) and cosx=(1-u^2)/(u^2+1) from tan(x/2). After that I can use that to plug in the integral sinx/(sinx+cosx) to make the equation solvable.
 
I don't quite get it. Just substitute [itex]u=\tan(x/2)[/itex] and you'll see that the equation is correct...
 
thanks, i got it now
 
Hello,

Another easier approach :

You have u=tan(x/2), draw the triangle.

Now, sin(x)=sin( 2 (x/2) ) = 2 sin(x/2) cos(x/2) , get sin(x/2) & cos(x/2) from your triangle.
cos(x)=cos( 2 (x/2) ) = [ cos(x/2) ]^2 - [ sin(x/2) ]^2
 

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