1. Feb 1, 2006

### nick727kcin

im stuck at this integral where u=t

i know the integral of (du)/(u^2+1) equals arctan(u), i think that might have something to do with it

thanks

2. Feb 1, 2006

### HallsofIvy

Staff Emeritus
But you don't HAVE "u2+ 1", you have "u2+ 3". So try $u^2+ 3= 3(\frac{1}{3}u^2+ 1)$ and let $v= \frac{1}{\sqrt{3}}u$. Then $dv= \frac{1}{\sqrt{3}}du$ so that $du= \sqrt{3}dv$. Of course, $u^2+ 3= 3(\frac{1}{3}u^2+1)= 3(v^2+ 1). When u= 1, [itex]v= \frac{1}{\sqrt{3}}$ and when $u= e^t$, $v= \frac{1}{3}e^t$. Your integral becomes:
$$\frac{\sqrt{3}}{3}\int_\frac{1}{\sqrt{3}}^{\frac{1}{\sqrt{3}}e^t}\frac{dv}{v^2+1}$$

Yes, the integral of that is $\frac{\sqrt{3}}{3}arctan(v)$ evaluated between the two limits.

Last edited: Feb 1, 2006
3. Feb 1, 2006

### nick727kcin

whoaa i dont think im that advanced yet, but thanks man. ill try to use it

4. Feb 1, 2006

### HallsofIvy

Staff Emeritus
"Advanced"?? The most advanced calculus used was the arctan integral. The rest is all algebra.

5. Feb 1, 2006

### nick727kcin

would this be equal to ((3^.5)/3) x ((2Theta)/6)