Figuring Out Integral for Nevermind I Seem to Have

  • Thread starter Thread starter roeb
  • Start date Start date
  • Tags Tags
    Integral
roeb
Messages
98
Reaction score
1
Nevermind, I seem to have figured it out

Homework Statement


show that:
\int \frac{dN}{N} = \int_{0}^{\theta_max} \frac{dcos \theta}{2} \frac{1 + \frac{v^2}{v_0^2} cos(2 \theta ) }{ \sqrt{ 1 - \frac{v^2 sin^2 \theta}{v_0^2} } }

Homework Equations


The Attempt at a Solution


I'm having a lot of difficulty doing this...
Note that sin(\theta_{max} ) = \frac{v_0}{v}
so after a bunch of algebra I get:

\int \frac{dN}{N} = cot(\theta_{max} ) \int_{0}^{\theta_{max} } \frac{2y^2 - 1}{\sqrt{y^2 - 1} }
I am fairly confident that is correct because I keep on getting it.
Unfortunately, I can't seem to integrate this at all.
 
Last edited:
Physics news on Phys.org
<br /> \int \frac{dN}{N} = cot(\theta_{max} ) \int_{0}^{\theta_{max} } \frac{2y^2 - 1}{\sqrt{y^2 - 1} }<br />

If you are finding it hard to integrate the right side, try u = sqrt(y^2-1) .. fairly simple to integrate
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top