By the substitution you argued 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: Integral of 1/sqrt(1-x^2)
However your original integral can likewise be solved by$x=\sinh t$ and $dx=\cosh t\, dt$ which gives$$\int \frac\cosh t\cosh t \, dt = \int 1\, dt=t=\operatornameargsinh x = \ln (x+\sqrtx^2+1)+C,$$since $\sqrt1+x^2=\sqrt1+\sinh^2 t=\cosh t$.
See hyperbolic functions and their inverses.
If girlfriend are acquainted (=used come manipulate) with the hyperbolic features then $x=a\sinh t$ is worth trying at any time you view the expression $\sqrta^2+x^2$ in your integral ($a$ gift an arbitrary constant).
edited Aug 5 "12 in ~ 14:29
answer Aug 5 "12 at 14:00
martin SleziakMartin Sleziak
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$\begingroup$ just how do you get from $\int \frac1\sqrt1+x^2 dx$ come $\int \frac1cosh tdx=\int \fraccosh tcosh tdt$? $\endgroup$
Aug 5 "12 at 14:27
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A variant of the hyperbolic duty substitution is to let $x=\frac12\left(t-\frac1t\right)$. Keep in mind that $1+x^2=\frac14\left(t^2+2+\frac1t^2\right)$.
So $\sqrt1+x^2=\frac12\left(t+\frac1t\right)$. The was the whole suggest of the substitution, that is a rationalizing substitution that makes the square root simple. Also, $dx=\frac12\left(1+\frac1t^2\right)\,dt$.
Carry out the substitution. "Miraculously," ours integral simplifies to $\int \fracdtt$.
answer Aug 5 "12 at 15:26
André NicolasAndré Nicolas
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Put $x=\tan y$, so the $dx=\sec^2y \ dy$ and $\sqrt1+x^2=\sec y$
$$\int \frac1\sqrt1+x^2 dx$$
$$= \int \frac\sec^2y \ dy\sec y$$
$$=\int \sec y\, dy$$
which evaluate to $\displaystyle\ln|\sec y+\tan y|+ C$ , applying the conventional formula whose proof is here and also $C$ is one indeterminate constant for any indefinite integral.
$$=\ln|\sqrt1+x^2+x| + C$$
We deserve to substitute $x$ with $a \sec y$ for $\sqrtx^2-a^2$, and with $a \sin y$ because that $\sqrta^2-x^2$
edited Aug 5 "12 in ~ 14:37
answer Aug 5 "12 in ~ 14:05
rap bhattacharjeelab bhattacharjee
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Let, $x = \tan\theta$
$dx = \sec^2\thetad\theta$
substitute, $x$, $dx$
$$A=\int\sec\theta\left(\frac\sec\theta + \tan\theta\sec\theta + \tan\theta\right)d\theta$$
$$A=\int\left(\frac\sec^2\theta + \sec\theta\tan\theta\sec\theta + \tan\theta\right)d\theta$$
Let, $(\sec\theta + \tan\theta) = u$
$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$
$$A=\ln\vert\sec\theta + \tan\theta\vert+c$$
$$A=\ln\vert\sqrt<>1+\tan^2\theta + \tan\theta\vert+c$$
$A=\ln\vert\sqrt<>1+x^2 + x\vert+c$, whereby $c$ is a constant
reply Aug 5 "12 at 17:37
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