By the substitution you said you get$$int frac12sqrtt(t-1) ,dt=int frac1sqrt4t^2-4t ,dt=int frac1sqrt(2t-1)^2-1 ,dt$$Now the substitution $u=2t-1$ seems reasonable.
You are watching: Integral 1/sqrt(1-x^2)
However your initial integral can likewise be addressed by$x=sinh t$ and also $dx=cosh t, dt$ which gives$$int fraccosh tcosh t , dt = int 1, dt=t=operatornamearcsinh x = ln (x+sqrtx^2+1)+C,$$since $sqrt1+x^2=sqrt1+sinh^2 t=cosh t$.
See hyperbolic functions and also their inverses.
If friend are familiar (=used to manipulate) through the hyperbolic features then $x=asinh t$ is worth trying anytime you check out the expression $sqrta^2+x^2$ in her integral ($a$ gift an arbitrarily constant).
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edited Sep 23 in ~ 7:00
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answer Aug 5 "12 at 14:00
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$egingroup$ just how do you acquire from $int frac1sqrt1+x^2 dx$ to $int frac1cosh tdx=int fraccosh tcosh tdt$? $endgroup$
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Aug 5 "12 at 14:27
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A variant of the hyperbolic function substitution is to let $x=frac12left(t-frac1t
ight)$. Note that $1+x^2=frac14left(t^2+2+frac1t^2
ight)$.
So $sqrt1+x^2=frac12left(t+frac1t ight)$. That was the whole allude of the substitution, it is a rationalizing substitution that renders the square root simple. Also, $dx=frac12left(1+frac1t^2 ight),dt$.
Carry out the substitution. "Miraculously," our integral simplifies come $int fracdtt$.
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answer Aug 5 "12 at 15:26
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Put $x= an y$, so that $dx=sec^2y dy$ and $sqrt1+x^2=sec y$
$$int frac1sqrt1+x^2 dx$$
$$= int fracsec^2y dysec y$$
$$=int sec y, dy$$
which evaluates to $displaystyleln|sec y+ an y|+ C$ , applying the typical formula whose proof is here and $C$ is one indeterminate constant for any type of indefinite integral.
$$=ln|sqrt1+x^2+x| + C$$
We deserve to substitute $x$ v $a sec y$ because that $sqrtx^2-a^2$, and also with $a sin y$ for $sqrta^2-x^2$
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edited Aug 5 "12 in ~ 14:37
reply Aug 5 "12 in ~ 14:05
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$$A=intfrac1sqrt<>1+x^2$$
Let, $x = an heta$
$dx = sec^2 hetad heta$
substitute, $x$, $dx$
$$A=intleft(frac1sec heta ight)sec^2 hetad heta$$
$$A=intsec hetad heta$$
$$A=intsec hetaleft(fracsec heta + an hetasec heta + an heta ight)d heta$$
$$A=intleft(fracsec^2 heta + sec heta an hetasec heta + an heta ight)d heta$$
Let, $(sec heta + an heta) = u$
$(sec^2 heta + sec heta an heta)d heta = du$
$$A=intfracduu$$
$$A=lnu+c$$
$$A=lnvertsec heta + an hetavert+c$$
$$A=lnvertsqrt<>1+ an^2 heta + an hetavert+c$$
$A=lnvertsqrt<>1+x^2 + xvert+c$, whereby $c$ is a constant
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reply Aug 5 "12 at 17:37
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