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


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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).


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edited Aug 5 "12 in ~ 14:29
answer Aug 5 "12 at 14:00
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martin SleziakMartin Sleziak
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$\begingroup$ just how do you get from $\int \frac1\sqrt1+x^2 dx$ come $\int \frac1cosh tdx=\int \fraccosh tcosh tdt$? $\endgroup$
–user2723
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 $\int \fracdtt$.


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answer Aug 5 "12 at 15:26
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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 $\sqrta^2-x^2$


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edited Aug 5 "12 in ~ 14:37
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rap 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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