Question

\[\int\frac{1}{\sqrt{1+x^{2}}}
dx\]


Without integration by trig. substitution

Answer 1

how about using integral formula?

Answer 2

Well...
Let \(x = \tan t\)
then, \(dx = \sec^2tdt\)
So, your integral becomes:
\(\int \sec t dt\)

which is back to @Callisto explanation in a post a few questions back.

WHEEEEEEEEEEE!!!!!!!!!

Answer 3

Apoor.... NO TRIGO!!!!

Answer 4

I am BLIND. >.<

Answer 5

PUT \[x=\tan \theta\]

Answer 6

\[1+\tan ^2(\theta)=\sec^2(\theta)\]

Answer 7

@nitz According to the question: Without integration by trig. substitution

Answer 8

sorry i didnt see it

Answer 9

The standard integrand formula is:
\(\large \int \frac 1 {\sqrt{a^2 +x^2}} = ln|x+\sqrt{x^2 + a^2}| + k\)

Now how do we derive this hmm...

Answer 10

Without trig substitution.
It's just a simple substitution and manipulation (:

Answer 11

isnt it \[\tan^{-1} (x)\]

Answer 12

no trig

Answer 13

You sure you do this WITHOUT that trig. thing?? Hmm..

Answer 14

It's workable

Answer 15

@nitz that's for integral of 1/(1+x^2)

Answer 16

i remember....its sumthing to do with hyperbolic functions.....

Answer 17

I suppose we can't let\[x=\sinh(u)?\]

Answer 18

No @Mimi_x3 that sub. didn't work, back to square one.

Answer 19

I don't know if this violates the trig substitution constraint but I'll post it anyway, maybe inspire some other thought.

Let \[x=\sinh(u)\]\[dx=\cosh(u)\]

So,\[\int\limits_{}\frac{1}{\sqrt{1+x^2}}=\int\limits_{}\frac{1}{\sqrt{1+sinh^2(u)}}cosh(u)=\int\limits_{}\frac{1}{\cosh(u)}\cosh(u)du=\int_{}du=u\]

Must put back in terms of x.
\[x=\sinh(u)\rightarrow u=\sinh^{-1}(x)\]

So,
\[\int\limits_{}\frac{1}{\sqrt{1+x^2}}dx=\sinh^{-1}(x)+c\]

Answer 20

Supposed to be a du after the second integral...