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

[Froskoy]

Homework Statement

\int \frac{1}{2t^2+4} \mathrm{d}t[\tex]<h2>Homework Equations</h2><br /> \int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c<h2>The Attempt at a Solution</h2><br /> Looks quite easy, but this is what&#039;s annoying me: the two methods below should be identical, but something&#039;s gone wrong and I can&#039;t work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren&#039;t they both the same?<br /> <br /> \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<br /> <br /> \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<br /> <br /> With very many thanks,<br /> <br /> Froskoy.

 

[kru_]

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

 

[Ray Vickson]

Homework Statement

\int \frac{1}{2t^2+4} \mathrm{d}t[\tex]<br /> <br /> <br /> <h2>Homework Equations</h2><br /> \int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c<br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> Looks quite easy, but this is what&#039;s annoying me: the two methods below should be identical, but something&#039;s gone wrong and I can&#039;t work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren&#039;t they both the same?<br /> <br /> \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<br /> <br /> \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<br /> <br /> With very many thanks,<br /> <br /> Froskoy.

<br /> <br /> You forgot to use \sqrt{2} dt in the first integral.<br /> <br /> RGV