Struggling with Trigonometric Substitution and Anti-Derivatives?

  • Thread starter Thread starter domesticbark
  • Start date Start date
  • Tags Tags
    Derivative
domesticbark
Messages
6
Reaction score
0
anti derive sqrt{1 + x^(-2/3)}


So this isn't actually a homework problem... just a problem that's been bugging me. Supposedly it's possible to do this using u substitution, but I'm having quite a bit of trouble... I've tried making U = x^(-2/3) and I realize that's probably not the best thing to do... Either way I've been able to get everything to a u and get du in there, it's just in comes down to a odd multiplication thing. I'm not looking for an answer really, perhaps somehow to be led in the right direction. Or even just show me how to anti derive sqrt {x^2 + x}. That would help a bit since I'm stuck in a similar scenario.
 
Physics news on Phys.org
Pick x^{-2/3} = \tan(\theta)^2, and then \sqrt{1 + x^{-2/3}} = \sqrt{1 + \tan(\theta)^2} = \sqrt{\sec(\theta)^2}. You'll have to be sure to find dx = \frac{d}{d\theta} \tan(\theta)^{-3} and further work for the substitution there, and later some integration by parts for what shows up afterwards. But I feel that approach should work.
 
Last edited:
The trick to using substitutions is to turn parts of the integrand into things that you already know how to integrate. You might also have to make multiple substitutions before you get something completely manageable. For instance,

\int \frac{du}{\sqrt{1+u^2}}

can be done by a substitution u=\tan v. In your case, you can manipulate your integrand to a form similar to this (though there's another factor of u to some power that might make things a bit harder.) I think that some substitutions of this form are helpful, but haven't worked the whole thing out myself.
 
You should try combining the terms under the root, see if that doesn't help a bit, at this point you should be looking for a substitution. Good luck.

Joe
 
So what you guys are saying is I can sub other things in for u? If \chi-2/3= U, can I really just make U = tan\Theta? I'm just a little confused since I'm semi new to calculus. And that makes things a little more complicated. how would I get d\Theta to replace du? Ahh I am confused. Either way, thank you for the help. I'm sure I'll figure it out eventually.
 
Last edited by a moderator:
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