just how you integrate $\\frac1\\sqrt1+x^2$ using complying with substitution? $1+x^2=t$ $\\Rightarrow$ $x=\\sqrtt-1 \\Rightarrow dx = \\fracdt2\\sqrtt-1dt$... Now I\"m stuck. I don\"t know exactly how to continue using substitution rule.

By the substitution you suggested you get$$\\int \\frac12\\sqrtt(t-1) \\,dt=\\int \\frac1\\sqrt4t^2-4t \\,dt=\\int \\frac1\\sqrt(2t-1)^2-1 \\,dt$$Now the substitution $u=2t-1$ seems reasonable.

You are watching: 1/sqrt(1-x^2)

However your initial integral can also be resolved by$x=\\sinh t$ and $dx=\\cosh t\\, dt$ i beg your pardon gives$$\\int \\frac\\cosh t\\cosh t \\, dt = \\int 1\\, dt=t=\\operatornamearcsinh x = \\ln (x+\\sqrtx^2+1)+C,$$since $\\sqrt1+x^2=\\sqrt1+\\sinh^2 t=\\cosh t$.

See hyperbolic functions and their inverses.

If friend are acquainted (=used come manipulate) through the hyperbolic attributes then $x=a\\sinh t$ is worth trying anytime you check out the expression $\\sqrta^2+x^2$ in your integral ($a$ being an arbitrary constant).

re-superstructure
point out
follow
edited Sep 23 in ~ 7:00

cutting board Andrews
answer Aug 5 \"12 in ~ 14:00

boy name SleziakMartin Sleziak
$\\endgroup$
6
$\\begingroup$ just how do you gain from $\\int \\frac1\\sqrt1+x^2 dx$ come $\\int \\frac1cosh tdx=\\int \\fraccosh tcosh tdt$? $\\endgroup$
–user2723
Aug 5 \"12 in ~ 14:27

| display 1 much more comment
13
$\\begingroup$
A variant of the hyperbolic role substitution is come let $x=\\frac12\\left(t-\\frac1t\\right)$. Keep in mind that $1+x^2=\\frac14\\left(t^2+2+\\frac1t^2\\right)$.

So $\\sqrt1+x^2=\\frac12\\left(t+\\frac1t\\right)$. That was the whole suggest of the substitution, the is a rationalizing substitution that renders the square source simple. Also, $dx=\\frac12\\left(1+\\frac1t^2\\right)\\,dt$.

Carry the end the substitution. \"Miraculously,\" our integral simplifies come $\\int \\fracdtt$.

re-superstructure
cite
monitor
reply Aug 5 \"12 in ~ 15:26

André NicolasAndré Nicolas
$\\endgroup$
5
$\\begingroup$
Put $x=\\tan y$, so that $dx=\\sec^2y \\ dy$ and $\\sqrt1+x^2=\\sec y$

$$\\int \\frac1\\sqrt1+x^2 dx$$

$$= \\int \\frac\\sec^2y \\ dy\\sec y$$

$$=\\int \\sec y\\, dy$$

which evaluates to $\\displaystyle\\ln|\\sec y+\\tan y|+ C$ , applying the standard formula whose proof is here and $C$ is an indeterminate constant for any indefinite integral.

$$=\\ln|\\sqrt1+x^2+x| + C$$

We can substitute $x$ with $a \\sec y$ for $\\sqrtx^2-a^2$, and with $a \\sin y$ because that $\\sqrta^2-x^2$

re-publishing
mention
monitor
edited Aug 5 \"12 at 14:37
answer Aug 5 \"12 in ~ 14:05

lab bhattacharjeelab bhattacharjee
1
$\\endgroup$
3
$\\begingroup$
$$A=\\int\\frac1\\sqrt<>1+x^2$$

Let, $x = \\tan\\theta$

$dx = \\sec^2\\thetad\\theta$

substitute, $x$, $dx$

$$A=\\int\\left(\\frac1\\sec\\theta\\right)\\sec^2\\thetad\\theta$$

$$A=\\int\\sec\\thetad\\theta$$

$$A=\\int\\sec\\theta\\left(\\frac\\sec\\theta + \\tan\\theta\\sec\\theta + \\tan\\theta\\right)d\\theta$$

$$A=\\int\\left(\\frac\\sec^2\\theta + \\sec\\theta\\tan\\theta\\sec\\theta + \\tan\\theta\\right)d\\theta$$

Let, $(\\sec\\theta + \\tan\\theta) = u$

$(\\sec^2\\theta + \\sec\\theta\\tan\\theta)d\\theta = du$

$$A=\\int\\fracduu$$

$$A=\\lnu+c$$

$$A=\\ln\\vert\\sec\\theta + \\tan\\theta\\vert+c$$

$$A=\\ln\\vert\\sqrt<>1+\\tan^2\\theta + \\tan\\theta\\vert+c$$

$A=\\ln\\vert\\sqrt<>1+x^2 + x\\vert+c$, whereby $c$ is a constant

share
mention
follow
answer Aug 5 \"12 in ~ 17:37
HOLYBIBLETHEHOLYBIBLETHE
$\\endgroup$
include a comment |

Thanks because that contributing response to carolannpeacock.comematics stack Exchange!

Please be certain to answer the question. Carry out details and also share her research!

But avoid

Asking because that help, clarification, or responding to various other answers.Making statements based upon opinion; ago them increase with referrals or an individual experience.

Use carolannpeacock.comJax to format equations. Carolannpeacock.comJax reference.

See more: D&Amp;D Puzzles That Require Teamwork, Vitamin D Deficiency: Symptoms & Treatment

Draft saved

send

### Post as a guest

surname
email Required, but never shown

### Post together a guest

surname
email

Required, however never shown

Featured top top Meta
Visit conversation
1
Solving the integral $\\int\\frac1\\sqrtx^2+1\\,dx$
4
How to integrate $\\int dx \\over \\sqrt1 + x^2$
6
The integral $\\int\\frac2(2y^2+1)(y^2+1)^0.5 dy$
1
Evaluate the integral $\\int \\frac\\cos(x)\\sqrt1+\\sin^2(x) \\, dx$
1
Compute $\\int_0^1\\frac \\sqrtx(x+3)\\sqrtx+3dx.$
associated
4
Integration that $\\int \\fracdxx^2\\sqrtx^2 + 9$ utilizing trigonometric substitution
1
integrate using substitution
3
just how to integrate $\\frac1x\\sqrt1+x^2$ making use of substitution?
0
Integration making use of hyperbolic substitution
0
resolve $\\int\\frac\\sqrtx-1xdx$ by making use of substitution
2
incorporate $x^2\\sin(2x)$ using $u$-substitution
1
just how to incorporate $\\int \\frac816-e^4x \\carolannpeacock.comrm dx$ making use of trigonometric substitution?
1
integrate using $t$ substitution
hot Network questions much more hot inquiries

concern feed