# How do I integrate this: 1 to 2∫ √(25+100t^2)?

1. Sep 29, 2011

### winkzy

I'm trying to find the arc length and I'm able to get this far. But it's been a long time since I've done calculus so I forgot.

1 to 2∫ √(25+100t^2)

I tried to do u = 25+100t^2, du = 200t dt, dt = 1/200 du, but it doesn't look right.

2. Sep 29, 2011

### mathman

dt=du/(200t), where t = √[(u-25)/100]

3. Sep 29, 2011

### HallsofIvy

Staff Emeritus
That's a pretty standard trig substitution: $sin^2(\theta)+ cos^2(\theta)= 1$. Divide through by $cos^2(\theta)$, $tan^2(\theta)+ 1= cos^2(\theta)$.

Therefore, let $5tan(\theta)= (1/10)t$ so that $tan^2(\theta)= (25/100)t^2$, $100t^2+ 25= 25tan^2(\theta)+ 25= 25(tan^2(\theta)+ 1)= 25cos^2(\theta)$ and so $\sqrt{100t^2+ 25}= 5cos(theta)$.

Of course, $d(tan(\theta))= sec^2(\theta)d\theta$ so that $(1/10)dt= 5sec^2(\theta)d\theta$.

4. Sep 29, 2011

### Char. Limit

I'm pretty sure that tan^2(x)+1 is not cos^2(x). I believe Halls of Ivy meant sec^2(x).

5. Sep 29, 2011

### sachav

Also, if you define $5\tan \theta = \frac{1}{10}t$, then $\tan^2 \theta = \frac{1}{2500}t^2$