What is the Integral of sin x * sqrt(1+cos2x)?

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Homework Help Overview

The problem involves evaluating the integral ∫(sin x)√(1+cos2x) dx, which is situated within the context of integral calculus and trigonometric identities.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of the integral, with one suggesting a substitution of y = cos(x). Others explore the transformation of the integral into a different form involving √(1+y^2) and consider trigonometric or hyperbolic substitutions.

Discussion Status

The discussion is active, with participants sharing their progress and seeking further guidance. Some have reached a point of evaluating -∫sec3v dv and are contemplating the use of integration by parts or alternative methods.

Contextual Notes

There are indications of confusion regarding the initial expression and the use of LaTeX formatting. Participants are navigating through various substitutions and transformations, reflecting the complexity of the integral.

wlooi
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Homework Statement



∫(sin x)√(1+cos2x) dx
 
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What you have written makes no sense (even if the latex were working). you have \sqrt{} with nothing inside the square root and have a subscript that makes no sense.

Did you mean [itex]\int sin(x)\sqrt{1+ cos^2(x)}dx[/itex]?

If so, start by letting y= cos(x).
 
Yes, that's exactly what I was try to make. Well, this is my first time using this so sorry..
 
Well, I got myself until:

[itex]\large -[/itex] [itex]\int \sqrt{1+y^2}[/itex] [itex]\large dy[/itex]

then I stucked again.
 
wlooi said:
Well, I got myself until:

[itex]\large -[/itex] [itex]\int \sqrt{1+y^2}[/itex] [itex]\large dy[/itex]

then I stucked again.

Try and think of a trig (or hyperbolic trig) substitution that will turn 1+y^2 into the square of something so you can get rid of the square root.
 
Argh... I don't seem to get it after cracking my head for a while. Can I have a little bit more tips?
 
Let's see, ...

1+tan2(θ) = sec2(θ)

1+sinh2(u) = cosh2(u)
 
using 1+tan2v=sec2v , I got it until :

-∫sec3v dv

am I suppose to do by parts or is there other ways?
 
wlooi said:
using 1+tan2v=sec2v , I got it until :

-∫sec3v dv

am I suppose to do by parts or is there other ways?

Parts. Split it into sec(v)^2*dv and sec(v).
 
  • #10
Okay, I think I got it. Thanks for the help from everyone.^^
 

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