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

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Homework Help Overview

The problem involves evaluating the integral \(\int_{0}^{\pi/4}\frac{\sin x}{\sqrt{\cos2x}}dx\), which includes trigonometric functions and a square root. The discussion centers around the challenges of integrating this expression, particularly due to the square root in the denominator and the behavior of the integrand at the upper limit of integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various substitution methods, including \(y = \tan(x/2)\) and rewriting \(\cos(2x)\). Some question whether algebraic errors might be present in the attempts. There is also a debate regarding the necessity of considering limits due to the behavior of the integral at the upper limit.

Discussion Status

The discussion is active, with participants offering different perspectives on the need for limits and the evaluation of the integral at specific points. Some guidance has been provided regarding the nature of the integral, including its classification as improper, but there is no explicit consensus on the evaluation methods or outcomes.

Contextual Notes

Participants note that the integral poses challenges due to the square root and the potential for undefined behavior at the upper limit. There is mention of the integral being improper, which adds complexity to the evaluation process.

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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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gop said:
[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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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.
 
Gib Z said:
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.)
 
@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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