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

[gop]

Homework Statement

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

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.

 

[dynamicsolo]

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

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...]

 

[Gib Z]

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.

 

[dynamicsolo]

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.)

 

[Rainbow Child]

@dynamicsolo I don't think he has to consider any limits, with your substitution the indefinite integral reads

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

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