Question

integrate root(x^(2)+1)
how to substitute here

Answer 1

$$\int\sqrt{x^2+1}\,dx$$Let \(x=\sinh t\) so \(dx=\cosh t\,dt\):$$\int\sqrt{x^2+1}dx=\int\cosh t\sqrt{\sinh^2 t+1}\,dt=\int\cosh^2 t\,dt$$

Answer 2

well, but if haven't study hyperbolic trigs any other way?

Answer 3

We know \(\cosh^2 t=\frac12(1+\cosh2t)\) so we have:$$\int\cosh^2t\,dt=\frac12\int(1+\cosh2t)\,dt=\frac12\left(t+\frac12\sinh2t\right)+C$$

Answer 4

Recall \(\sinh 2t=2\sinh t\cosh t\) so:$$\DeclareMathOperator{\arcsinh}{arcsinh}

\frac12 t+\frac14(2\sinh t\cosh t)+C=\frac12t+\frac12\sinh t\cosh t+C$$Observe we have \(x=\sinh t\Leftrightarrow \arcsinh x=t\) and \(\cosh t=\sqrt{1+\sinh^2t}=\sqrt{1+x^2}\):$$\frac12\arcsinh x+\frac12x\sqrt{1+x^2}+C$$

Answer 5

If you don't use hyperbolic substitutions it'll be uglier... but you could observe that since \(1+\tan^2 t=\sec^2t\) we can substitute \(x=\tan t\)

Answer 6

@syltan x = tan t seems good to me. Can you use it?

Answer 7

x=tan^2 this looks more familiar to me thank you.

Answer 8

@Loser66 you'll end up having to integrate \(\sec^3 t\) though :-/

Answer 9

i did it not long time ago just copy paste.

Answer 10

yes, but it's easier.

Answer 11

and we are familiar with it.

Answer 12

\(\sec^3t\) is much uglier! why use parts when you can avoid it all together... :-(

Answer 13

hahaha.... ok, I hand off!! you convinced me that hyperbolic is more beautiful than normal trigs. You won!!

Answer 14

Well... \(\cosh x,\sinh x\) are an eigenbasis for \(\dfrac{d^2}{dx^2}\)... just look how nice:$$y''=y\\\frac{d}{dx}\sinh x=\cosh x\\\frac{d}{dx}\cosh x=\sinh x\\\cosh^2x-\sinh^2 x=0$$

Answer 15

hey, the third one is not = -sinh x?

Answer 16

Nope! no pesky sign changes! much nicer

Answer 17

As I told you, I knew it but a little only. I look them up at my note, nothing there, but definition. So, although i made question and you answered that no pesky sign changes, I have no way to confirm.

Answer 18

@Loser66 if \(\frac{d}{dx}\cosh x=-\sinh x\) then surely it would not satisfy \(y''=y\)? :-)

Answer 19

Hyperbolic functions help you stress less in situations where subbing is difficult. They work the same way as normal trig subs wherever trig identities are concerned, with a slightly difficult end result to derive (that takes memorization i'd think)