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-   -   Integrating 1/(Z^2 + A^2) - something wrong? (http://www.physicsforums.com/showthread.php?t=560829)

 Froskoy Dec16-11 01:51 PM

Integrating 1/(Z^2 + A^2) - something wrong?

1. The problem statement, all variables and given/known data
$$\int \frac{1}{2t^2+4} \mathrm{d}t[\tex] 2. Relevant equations [tex]\int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c$$

3. The attempt at a solution
Looks quite easy, but this is what's annoying me: the two methods below should be identical, but something's gone wrong and I can't work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren't they both the same?

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c$$

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c$$

With very many thanks,

Froskoy.

 kru_ Dec16-11 01:55 PM

Re: Integrating 1/(Z^2 + A^2) - something wrong?

Are you sure that c represents the same value in each?

 Ray Vickson Dec16-11 02:37 PM

Re: Integrating 1/(Z^2 + A^2) - something wrong?

Quote:
 Quote by Froskoy (Post 3671902) 1. The problem statement, all variables and given/known data $$\int \frac{1}{2t^2+4} \mathrm{d}t[\tex] 2. Relevant equations [tex]\int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c$$ 3. The attempt at a solution Looks quite easy, but this is what's annoying me: the two methods below should be identical, but something's gone wrong and I can't work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren't they both the same? $$\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c$$ $$\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c$$ With very many thanks, Froskoy.
You forgot to use $\sqrt{2} dt$ in the first integral.

RGV

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