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Integral with sin(x), cos(2x), and sqrt

  • Thread starter gop
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  • #1
gop
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Homework Statement



[tex]\int_{0}^{\pi/4}\frac{\sin x}{\sqrt{\cos2x}}dx[/tex]

Homework Equations





The Attempt at a Solution



I used the standard substitutions (like y = tan(x/2)) and the substitution to get rid of the square root). I also tried to rewrite cos(2x) in many different ways but in the end I always end up with more square roots and/or power terms of which I can't compute the integral.
 
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Answers and Replies

  • #2
dynamicsolo
Homework Helper
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[tex]\int_{0}^{\pi/4}\frac{\sin x}{\sqrt{\cos2x}}dx[/tex]
Maybe you ought to show something you tried: perhaps there's just an algebra error somewhere?

You wrote cos 2x as 2(cos x)^2 - 1 , used u = cos x , and that didn't work?

[EDIT: Whoa-hey! I just spotted something else: this is a Type II improper integral. The denominator is zero at the upper limit of the integration. I think the indefinite integral works OK, but you'll want to use a limit at the (pi)/4 end to see whether this thing actually converges or not...]
 
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  • #3
Gib Z
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No need to use a limit (unless your doing this in the formal sense), the value just plugs straight in. For the value of the indefinite integral at pi/4, i got 0.
 
  • #4
dynamicsolo
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No need to use a limit (unless your doing this in the formal sense), the value just plugs straight in. For the value of the indefinite integral at pi/4, i got 0.
As for the behavior at the vertical asymptote, I agree that the limit turns out to be unnecessary; however, I do not agree that the value at pi/4 is zero... (This improper integral does converge all right.)
 
  • #5
@dynamicsolo I don't think he has to consider any limits, with your substitution the indefinite integral reads

[tex]I=-\frac{\sqrt{2}}{2}\,\ln\left(\sqrt{2}\,\cos x+\sqrt{\cos(2\,x)}\right)[/tex]

which is well defined in [itex][0,\frac{\pi}{4}][/itex]
 

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