Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule.

share|improve this question
Are you sure this is a good idea to proceed? –  Siminore Aug 5 '12 at 13:53
$1+x^2=t\implies 2xdx=dt\implies dx=\frac{dt}{2x}$ –  Belgi Aug 5 '12 at 13:54
@Belgi that's the same results alvoutilla found, just written in different letters. –  Kevin Carlson Aug 5 '12 at 13:59
Is it a typo that you have $dt$ twice in the expression for $dx$? –  Martin Sleziak Aug 5 '12 at 14:02
@KevinCarlson - indeed, but I wanted to show to the PO that he doesn't need to first 'solve' for $x$ to get that –  Belgi Aug 5 '12 at 14:11

4 Answers 4

up vote 10 down vote accepted

By the substitution you suggested you get $$ \int \frac1{2\sqrt{t(t-1)}} \,dt= \int \frac1{\sqrt{4t^2-4t}} \,dt= \int \frac1{\sqrt{(2t-1)^2-1}} \,dt $$ Now the substitution $u=2t-1$ seems reasonable.

However your original integral can also be solved by $x=\sinh t$ and $dx=\cosh t\, dt$ which gives $$\int \frac{\cosh t}{\cosh t} \, dt = \int 1\, dt=t=\operatorname{argsinh} x = \ln (x+\sqrt{x^2+1})+C,$$ since $\sqrt{1+x^2}=\sqrt{1+\sinh^2 t}=\cosh t$.

See hyperbolic functions and their inverses.

If you are familiar (=used to manipulate) with the hyperbolic functions then $x=a\sinh t$ is worth trying whenever you see the expression $\sqrt{a^2+x^2}$ in your integral ($a$ being an arbitrary constant).

share|improve this answer
How do you get from $\int \frac{1}{\sqrt{1+x^2}} dx$ to $\int \frac{1}{cosh t}dx=\int \frac{cosh t}{cosh t}dt$? –  alvoutila Aug 5 '12 at 14:27
@alvoutila $\sqrt{1+\sinh^2 t}=\cosh t$. (I've added this to my post, too.) –  Martin Sleziak Aug 5 '12 at 14:30
Instead of using hyperbolic functions, if you use normal trig functions (e.g. x = sin(t)), then you end up with integral = arcsin(x). Why is this incorrect? EDIT: NEVERMIND... 1 + sin**2 =/= cos –  user60462 yesterday
@user60462 As shown in other answers, you can use trigonometric substitution with $x=\tan t$. –  Martin Sleziak yesterday

Put $x=\tan y$, so that $dx=\sec^2y \ dy$ and $\sqrt{1+x^2}=\sec y$

$$\int \frac{1}{\sqrt{1+x^2}} dx$$

$$= \int \frac{\sec^2y \ dy}{\sec y}$$

$$=\int \sec y\, dy$$

which evaluates to $\displaystyle\ln|\sec y+\tan y|+ C$ , applying the standard formula whose proof is here and $C$ is an indeterminate constant for any indefinite integral.

$$=\ln|\sqrt{1+x^2}+x| + C$$

We can substitute $x$ with $a \sec y$ for $\sqrt{x^2-a^2}$, and with $a \sin y$ for $\sqrt{a^2-x^2}$

share|improve this answer

A variant of the hyperbolic function substitution is to let $x=\frac{1}{2}\left(t-\frac{1}{t}\right)$. Note that $1+x^2=\frac{1}{4}\left(t^2+2+\frac{1}{t^2}\right)$.

So $\sqrt{1+x^2}=\frac{1}{2}\left(t+\frac{1}{t}\right)$. That was the whole point of the substitution, it is a rationalizing substitution that makes the square root simple. Also, $dx=\frac{1}{2}\left(1+\frac{1}{t^2}\right)\,dt$.

Carry out the substitution. "Miraculously," our integral simplifies to $\int \frac{dt}{t}$.

share|improve this answer


Let, $x = \tan\theta$

$dx = \sec^{2}\theta{d\theta}$

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, $(\sec\theta + \tan\theta) = u$

$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$



$$A=\ln{\vert\sec\theta + \tan\theta\vert}+c$$

$$A=\ln{\vert\sqrt[]{1+\tan^2\theta} + \tan\theta\vert}+c$$

$A=\ln{\vert\sqrt[]{1+x^2} + x\vert}+c$, where $c$ is a constant

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.