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

[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.

 

[mathman]

dt=du/(200t), where t = √[(u-25)/100]

 

[HallsofIvy]

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.

 

[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).

 

[sachav]

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