Solve Tricky Integral: \int \sqrt{\theta+\frac{1}{2}\theta^{-1/2}}d\theta

  • Thread starter Thread starter Stratosphere
  • Start date Start date
  • Tags Tags
    Integral
Stratosphere
Messages
373
Reaction score
0

Homework Statement


I can't seem to figure out what I need to do in order to solve this one.
\int \sqrt{\theta+\frac{1}{2}\theta^{-1/2}}d\theta


Homework Equations





The Attempt at a Solution



I haven't the slightest clue as to what technique to use, it doesn't look to be a u substitution problem though (I could be wrong).
I typed it in on an online calculator and I got some really strange expression.

http://integrals.wolfram.com/index.jsp?expr=%28x%2B1%2F2x^%28-1%2F2%29%29^%281%2F2%29&random=false
 
Last edited:
Physics news on Phys.org
It is a substitution problem. Since you already cheated and looked at wolfram you may as well use that cheat. Look at the argument of the ArcSinh. That suggests if you substitute u=sqrt(2)x^(3/4) you should be able to reduce it to something you can handle. In fact, once you hack through the radicals you should be able to show that aside from the numeric factors, it becomes sqrt(u^2+1)du. Can you integrate that?
 
I don't want to use wolfram, I just used it because I couldn't figure it out. I didn't cheat on homework or anything I am self learning it. So what would be the first step starting from the original equation.
 
Well, just glancing at the integral it seems like the substitution u=\sqrt{\theta} is a decent place to start...have you tried that? If so, how far did you get with it?
 
Stratosphere said:
I don't want to use wolfram, I just used it because I couldn't figure it out. I didn't cheat on homework or anything I am self learning it. So what would be the first step starting from the original equation.

I'm exaggerating on the 'cheating' aspect. But I already suggested an initial substitution in my last post.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top