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).

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edited Sep 23 in ~ 7:00 cutting board Andrews
answer Aug 5 \"12 in ~ 14:00 boy name SleziakMartin Sleziak
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$\\begingroup$ 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

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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 $\\int \\fracdtt$.

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reply Aug 5 \"12 in ~ 15:26 André NicolasAndré Nicolas
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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 $\\sqrta^2-x^2$

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edited Aug 5 \"12 at 14:37
answer Aug 5 \"12 in ~ 14:05 lab bhattacharjeelab bhattacharjee
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$$A=\\int\\frac1\\sqrt<>1+x^2$$

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

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answer Aug 5 \"12 in ~ 17:37
HOLYBIBLETHEHOLYBIBLETHE
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