integral of <||cos t, sin t, t||> dt

the answer in the back has a natural log and i don't get it when i work it out???

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- Jun 12th 2008, 07:03 AMchris25integration of 3 variables
integral of <||cos t, sin t, t||> dt

the answer in the back has a natural log and i don't get it when i work it out??? - Jun 12th 2008, 07:19 AMMoo
- Jun 12th 2008, 07:45 AMchris25
no it means the length

- Jun 12th 2008, 07:47 AMMoo
- Jun 12th 2008, 07:56 AMchris25
evaluate the indefinite integral...

S||(cos t**i**+ sin t**j**+**t**k )|| dt - Jun 12th 2008, 07:58 AMMoo
- Jun 12th 2008, 08:07 AMMathstud28
$\displaystyle \int\sqrt{1+x^2}dx$

let $\displaystyle x=\tan(\theta)$

so $\displaystyle dx=\sec^2(\theta)d\theta$

Giving us after simplification

$\displaystyle \int\sec^3(\theta)d\theta$

Which by a manipulation and double parts gives http://www.mathhelpforum.com/math-he...-integral.html

$\displaystyle \frac{1}{2}\bigg[\ln|\sec(\theta)+\tan(\theta)|+\tan(\theta)\sec(\t heta)\bigg]$

Now back subbing in that $\displaystyle \theta=\arctan(x)$

we get

$\displaystyle \frac{1}{2}\bigg[\ln|\sqrt{x^2+1}+x|+x\sqrt{x^2+1}\bigg]$ - Jun 12th 2008, 08:40 PMmr fantastic
- Jun 12th 2008, 08:47 PMMathstud28
Yeah, but I wanted to supply a more commonly known sub

But to explain what Mr. F astutely noted

Let $\displaystyle x=\sinh(\theta)$

So $\displaystyle dx=\cosh(\theta)$

Giving us

$\displaystyle \int\sqrt{1+\sinh^2(\theta)}\cosh(\theta)d\theta=\ int\cosh^2(\theta)d\theta$

Now making the appropriate hyperbolic identity we get

$\displaystyle \frac{1}{2}\bigg[\theta+\cosh(\theta)\sinh(\theta)\bigg]$

Now just back sub - Jun 12th 2008, 11:32 PMMoo