# integration of 3 variables

• Jun 12th 2008, 07:03 AM
chris25
integration 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 AM
Moo
Hello,

Quote:

Originally Posted by chris25
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???

Is it ||...|| for the norm in basis 2 ?

It would be $\displaystyle ||\cos t, \sin t, t||=\sqrt{\cos^2t+\sin^2t+t^2}=\sqrt{1+t^2}$
• Jun 12th 2008, 07:45 AM
chris25
no it means the length
• Jun 12th 2008, 07:47 AM
Moo
Quote:

Originally Posted by chris25
no it means the length

Well, it has to be the same...

Can you state and write the exact question ?
Are there really < and > ?
• Jun 12th 2008, 07:56 AM
chris25
evaluate the indefinite integral...

S||(cos t i + sin t j + t k )|| dt
• Jun 12th 2008, 07:58 AM
Moo
Quote:

Originally Posted by chris25
evaluate the indefinite integral...

S||(cos t i + sin t j + t k )|| dt

So yeah, you have to calculate $\displaystyle \int \sqrt{1+t^2} dt$

But I'm not sure how to calculate it...
• Jun 12th 2008, 08:07 AM
Mathstud28
Quote:

Originally Posted by Moo
So yeah, you have to calculate $\displaystyle \int \sqrt{1+t^2} dt$

But I'm not sure how to calculate it...

$\displaystyle \int\sqrt{1+x^2}dx$

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 PM
mr fantastic
Quote:

Originally Posted by Mathstud28
$\displaystyle \int\sqrt{1+x^2}dx$

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

$\displaystyle \frac{1}{2}\bigg[\ln|\sqrt{x^2+1}+x|+x\sqrt{x^2+1}\bigg]$

The substitution $\displaystyle x = \sinh \theta$ will reduce the algebra.
• Jun 12th 2008, 08:47 PM
Mathstud28
Quote:

Originally Posted by mr fantastic
The substitution $\displaystyle x = \sinh \theta$ will reduce the algebra.

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 PM
Moo
Quote:

Originally Posted by mr fantastic
The substitution $\displaystyle x = \sinh \theta$ will reduce the algebra.

Actually, the OP mentioned that the answer contained a logarithm (Rofl)