Integral of sech x: Typo Error Check

  • Thread starter Thread starter DryRun
  • Start date Start date
  • Tags Tags
    Integral
DryRun
Gold Member
Messages
837
Reaction score
4
Homework Statement
I believe there might be a mistake in this statement from my online notes. A typo error in the last line, maybe?
http://s1.ipicture.ru/uploads/20111209/K6L7AWhS.jpgThe attempt at a solution
In the last line, from what i worked out, it should be (1/2)tan^(-1)(tanh... instead of 2tan^(-1)(tanh... Correct?
 
Physics news on Phys.org
Nope it is right. I would make two substitutions instead of one just because it is easier for me to think through that way. I would substitute u=(x/2) and then s=tanh(u) yielding the simple integral 2∫1/(s^2+1)ds=2tan^-1(s)=2tan^-1(tanh(u))=2tan^-1(tanh(x/2))+C

edit: The du for your equation would be .5sech^2(x/2)dx so you would pull a 2 out in front of the integral.
 
Thank you, PCSL. I see it now.
 
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...

Similar threads

Back
Top