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