# How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution?

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.

• 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

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).

• How do you get from $\int \frac{1}{\sqrt{1+x^2}} dx$ to $\int \frac{1}{cosh t}dx=\int \frac{cosh t}{cosh t}dt$? – user2723 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 Feb 26 '15 at 6:58
• @user60462 As shown in other answers, you can use trigonometric substitution with $x=\tan t$. – Martin Sleziak Feb 26 '15 at 10:39

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 $\int \frac{dt}{t}$.

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 $\sqrt{a^2-x^2}$

$$A=\int\frac{1}{\sqrt[]{1+x^2}}$$

Let, $x = \tan\theta$

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

substitute, $x$, $dx$

$$A=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}$$

$$A=\int{\sec\theta{d\theta}}$$

$$A=\int{\sec\theta\left(\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$

$$A=\int{\left(\frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$

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

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

$$A=\int\frac{du}{u}$$

$$A=\ln{u}+c$$

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

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

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