Homework Help: Can I do this? (integral)

1. Feb 5, 2005

Hello all, and sorry for making all those threads :shy:

I just want to know if I can do this (especially the last part)

$$\int{} \frac{-k\ \lambda \ dx \ x \vec{i} + 2\ k\ \lambda \dx \ \vec{j}}{(x^2 +4)^{3/2}}$$

= $$\int{} \frac{-k\lambda (x \vec{i} - 2\vec{j}) \ dx}{(x^2 +4)^{3/2}}$$
=
$$k\lambda \int \frac{(x\vec{i} - 2\vec{j})dx}{(x^2 +4)^{3/2}}$$
=$$k\lambda (\vec{i} - \vec{2j}) \int \frac{xdx}{(x^2 +4)^{3/2}}$$

I'm new to integrals.. is what I did okay? I think the last part is where it's wrong, but I can't figure how to let the "x" in the integral without doing this (because x is a variable)
$$\lambda \ and \ k$$ are constants

Last edited: Feb 5, 2005
2. Feb 5, 2005

dextercioby

No,i would advise you to evaluate each integral in the scalar form (i.e. to leave out the the unit vectors) and pay attention to every little detail.

Daniel.

P.S.You last integral can be done via a substitution.Do you know this method??

3. Feb 5, 2005

If by substitution you mean posing u=.. then replacing it, yes, we saw this in class.
I think I'll do what you said, since it seems like the only method (doing them in the scalar form). I tried to do them this way so I didnt have to take in account that it was positive/negative.
Thanks a lot for all the help, in all the threads and questions, dextercioby

4. Feb 5, 2005

dextercioby

Yes,ty.You made a mistake when taking out unit vectors outta the integral (namely when factoring).That's why is advised you to do them the normal way.

What do you know about (trigonometric) hyperbolic functions...?

Daniel.

5. Feb 5, 2005

Curious3141

I concur with daniel's advice about evaluating the scalar parts separately (and carefully !)

But you don't need a substitution to get the last integral, which can be more easily evaluating by observing the relationship to the differential of $$(x^2 + k)^{-\frac{1}{2}}$$, where $k$ is a constant (4 in this case).

6. Feb 5, 2005

dextercioby

$$\pm \int \frac{dx}{(x^{2}+4)^{\frac{3}{2}}}$$

Besides the obvious "sinh",could he find a simpler way...?

Daniel.

7. Feb 5, 2005

Curious3141

Yeah, but this wasn't that. :tongue2:

Have to be able to spot a form like $$g'(x)f(g(x))$$ instantly when you get adept at calculus.

8. Feb 5, 2005

dextercioby

$$f(g(x)) g'(x) dx=f(g(x)) dg(x)=f(u) du$$

where i made a very unobvious and spooky substitution ??? :tongue2: :tongue:

Daniel.

9. Feb 5, 2005

Curious3141

That IS spooky, but you could just as easily have spotted the form to begin with. :rofl: