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).

share

point out

follow

edited Aug 5 "12 in ~ 14:29

answer Aug 5 "12 at 14:00

martin SleziakMartin Sleziak

51.8k1818 gold badges162162 silver badges321321 bronze title

$\endgroup$

6

$\begingroup$ 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

| present 1 much more comment

13

$\begingroup$

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 $\int \fracdtt$.

re-publishing

point out

follow

answer Aug 5 "12 at 15:26

André NicolasAndré Nicolas

484k4343 gold badges504504 silver badges929929 bronze badges

$\endgroup$

add a comment |

5

$\begingroup$

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 $\sqrta^2-x^2$

share

mention

monitor

edited Aug 5 "12 in ~ 14:37

answer Aug 5 "12 in ~ 14:05

rap bhattacharjeelab bhattacharjee

266k1717 gold badges192192 silver badges305305 bronze title

$\endgroup$

include a comment |

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

re-superstructure

point out

monitor

reply Aug 5 "12 at 17:37

HOLYBIBLETHEHOLYBIBLETHE

2,63277 gold badges2626 silver- badges4545 bronze badges

$\endgroup$

include a comment |

## her Answer

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

Please be sure to*answer the question*. Provide details and also share your research!

But *avoid* …

Use allisonbrookephotography.comJax to layout equations. Allisonbrookephotography.comJax reference.

To learn more, check out our tips on writing an excellent answers.

See more: Aluminum Wire Size For 100 Amp Sub Panel 200 Feet Away, Questions 11

Draft saved

Draft discarded

### Sign increase or log in in

authorize up using Google

authorize up using Facebook

sign up using Email and Password

send

### Post together a guest

name

email Required, yet never shown

### Post together a guest

name

Required, however never shown

short article Your answer Discard

By clicking “Post your Answer”, girlfriend agree come our terms of service, privacy policy and also cookie policy

Featured ~ above Meta

Linked

1

Solving the integral $\int\frac1\sqrtx^2+1\,dx$

4

How to incorporate $\int dx \over \sqrt1 + x^2$

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 $\int_0^1\frac \sqrtx(x+3)\sqrtx+3dx.$

connected

4

Integration of $ \int \fracdxx^2\sqrtx^2 + 9 $ making use of trigonometric substitution

1

combine using substitution

3

how to combine $\frac1x\sqrt1+x^2$ making use of substitution?

0

Integration making use of hyperbolic substitution

0

deal with $ \int\frac\sqrtx-1xdx$ by making use of substitution

2

combine $x^2\sin(2x)$ using $u$-substitution

1

how to incorporate $\int \frac816-e^4x \allisonbrookephotography.comrm dx$ using trigonometric substitution?

1

combine using $t$ substitution

warm Network concerns more hot inquiries

concern feed

subscribe to RSS

concern feed To subscribe to this RSS feed, copy and also paste this URL right into your RSS reader.

allisonbrookephotography.comematics

agency

ridge Exchange Network

site design / logo design © 2021 ridge Exchange Inc; user contributions licensed under cc by-sa. Rev2021.9.22.40267

allisonbrookephotography.comematics ridge Exchange works finest with JavaScript allowed

her privacy

By clicking “Accept all cookies”, girlfriend agree ridge Exchange have the right to store cookie on your maker and disclose details in accordance with our Cookie Policy.