how you incorporate $frac1sqrt1+x^2$ using following substitution? $1+x^2=t$ $Rightarrow$ $x=sqrtt-1 Rightarrow dx = fracdt2sqrtt-1dt$... Now I"m stuck. Ns don"t know how to proceed using substitution rule.


*

By a substitution you suggested you get$$int frac12sqrtt(t-1) ,dt=int frac1sqrt4t^2-4t ,dt=int frac1sqrt(2t-1)^2-1 ,dt$$Now ns substitution $u=2t-1$ seems reasonable.

Você está assistindo: Integral de 1/raiz de x

However her original integral can additionally be addressed by$x=sinh t$ e $dx=cosh t, dt$ i beg your pardon 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 their inverses.

If you estão familiar (=used to manipulate) with the hyperbolic features then $x=asinh t$ is worth trying whenever friend see the expression $sqrta^2+x^2$ in your integral ($a$ being an arbitrarily constant).


share
point out
monitor
edited Sep 23 in ~ 7:00
*

thomas Andrews
160k1717 yellow badges187187 silver- badges359359 cobre badges
answered Aug 5 "12 in ~ 14:00
*

martinho SleziakMartin Sleziak
50.7k1818 gold badges163163 silver badges328328 cobre badges
$endgroup$
6
$egingroup$ How a partir de you get from $int frac1sqrt1+x^2 dx$ to $int frac1cosh tdx=int fraccosh tcosh tdt$? $endgroup$
–user2723
Aug 5 "12 at 14:27


| show 1 much more comment
13
$egingroup$
A different of a hyperbolic role substitution is come let $x=frac12left(t-frac1t ight)$. Keep in mind that $1+x^2=frac14left(t^2+2+frac1t^2 ight)$.

So $sqrt1+x^2=frac12left(t+frac1t ight)$. That was the whole point of a substitution, it is naquela rationalizing substitution that makes the quadrado root simple. Also, $dx=frac12left(1+frac1t^2 ight),dt$.

Carry fora the substitution. "Miraculously," our integrante simplifies come $int fracdtt$.


re-publishing
cite
monitor
reply Aug 5 "12 at 15:26
*

andré NicolasAndré Nicolas
486k4343 yellow badges506506 silver badges933933 bronze badges
$endgroup$
Add a comment |
5
$egingroup$
Put $x= an y$, so that $dx=sec^2y dy$ e $sqrt1+x^2=sec y$

$$int frac1sqrt1+x^2 dx$$

$$= int fracsec^2y dysec y$$

$$=int sec y, dy$$

which evaluate to $displaystyleln|sec y+ an y|+ C$ , applying the standard formula who proof is here e $C$ is an indeterminate constant ao any unknown integral.

$$=ln|sqrt1+x^2+x| + C$$

We deserve to substitute $x$ with $a sec y$ para $sqrtx^2-a^2$, and with $a sin y$ para $sqrta^2-x^2$


re-publishing
cite
monitor
edited Aug 5 "12 in ~ 14:37
reply Aug 5 "12 in ~ 14:05
*

lab bhattacharjeelab bhattacharjee
267k1717 gold badges193193 silver- badges307307 cobre badges
$endgroup$
Add naquela comment |
3
$egingroup$
$$A=intfrac1sqrt<>1+x^2$$

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, $(sec heta + an heta) = u$

$(sec^2 heta + sec heta an heta)d heta = du$

$$A=intfracduu$$

$$A=lnu+c$$

$$A=lnvertsec heta + an hetavert+c$$

$$A=lnvertsqrt<>1+ an^2 heta + an hetavert+c$$

$A=lnvertsqrt<>1+x^2 + xvert+c$, wherein $c$ is a constant


re-publishing
point out
follow
reply Aug 5 "12 at 17:37
HOLYBIBLETHEHOLYBIBLETHE
2,64277 gold badges2626 silver- badges4545 bronze badges
$endgroup$
Add der comment |

her Answer


Thanks porque o contributing response to classifimoveis.comematics Stack Exchange!

Please be sure to answer the question. Carry out details e share your research!

But avoid

Asking porque o help, clarification, or responding to various other answers.Making statements based upon opinion; back them up com references or an individual experience.

Use classifimoveis.comJax to style equations. classifimoveis.comJax reference.

To find out more, watch our advice on writing good answers.

Ver mais: " How I Met Your Mother 7 Temporada Download, How I Met Your Mother 7ª Temporada Torrent


Draft saved
Draft discarded

Sign up or log in in


authorize up using google
authorize up using Facebook
sign up making use of Email e Password
send

Post as naquela guest


name
email Required, however never shown


Post as naquela guest


surname
email

Required, yet never shown


article Your answer Discard

By click “Post her Answer”, girlfriend agree to our terms of service, privacy policy e cookie plan


Featured on portões
Visit bate-papo
Linked
1
Solving the integrante $intfrac1sqrtx^2+1,dx$
4
How to incorporate $int dx over sqrt1 + x^2$
6
The integral $intfrac2(2y^2+1)(y^2+1)^0.5 dy$
1
Evaluate the integrante $int fraccos(x)sqrt1+sin^2(x) , dx$
1
Compute $int_0^1frac sqrtx(x+3)sqrtx+3dx.$
related
4
Integration of crédito int fracdxx^2sqrtx^2 + 9 crédito using trigonometric substitution
1
integrate using substitution
3
how to incorporate $frac1xsqrt1+x^2$ making use of substitution?
0
Integration utilizing hyperbolic substitution
0
Solve crédito intfracsqrtx-1xdx$ by utilizing substitution
2
combine $x^2sin(2x)$ using $u$-substitution
1
como as to combine $int frac816-e^4x classifimoveis.comrm dx$ using trigonometric substitution?
9
Integrate via substitution: $x^2sqrtx^2+1;dx$
quente Network inquiries more quente questions

Question alimentação
subscribe to RSS
Question alimentação To i ordered it to this RSS feed, copy and paste this URL into your RSS reader.


classifimoveis.comematics
empresa
stack Exchange network
site esboço, projeto / em breve © 2021 stack Exchange Inc; user contributions licensed under cc by-sa. Rev2021.11.3.40639


classifimoveis.comematics stack Exchange functions best com JavaScript allowed
*

her privacy

By clicking “Accept tudo cookies”, friend agree ridge Exchange can store cookies on her device and disclose details in accordance com our Cookie Policy.