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.

So I came across the integral of $$\frac{1}{\sqrt{x^2+1}}$$ and I've tried substitution and it isn't on the list of common integrals. I used Wolfram Alpha, but it said $\sinh x$, but I was wondering if there is an alternate answer?

share|improve this question
1  
Shouldn't it be sinh^{-1}x? –  Amzoti Dec 12 '12 at 3:31

4 Answers 4

up vote 2 down vote accepted

Two substitutions suggest themselves

$x=\sinh(u)\Rightarrow\sqrt{1+x^2}=\cosh(u)\text{ and }\mathrm{d}x=\cosh(u)\,\mathrm{d}u$ which yields $$ \begin{align} \int\frac{\mathrm{d}x}{\sqrt{1+x^2}} &=\int\frac{\cosh(u)\,\mathrm{d}u}{\cosh(u)}\\ &=\int\,\mathrm{d}u \end{align} $$ $x=\tan(v)\Rightarrow\sqrt{1+x^2}=\sec(v)\text{ and }\mathrm{d}x=\sec^2(v)\,\mathrm{d}v$ which yields $$ \begin{align} \int\frac{\mathrm{d}x}{\sqrt{1+x^2}} &=\int\frac{\sec^2(v)\,\mathrm{d}v}{\sec(v)}\\ &=\int\frac{\mathrm{d}\sin(v)}{1-\sin^2(v)}\\ &=\frac12\int\frac{\mathrm{d}\sin(v)}{1-\sin(v)}+\frac12\int\frac{\mathrm{d}\sin(v)}{1+\sin(v)} \end{align} $$

share|improve this answer

It should be $\sinh^{-1} x$, which can also be written as $\ln (1+\sqrt{1+x^2})$ as shown in Wikipedia.

share|improve this answer

You can some times find antiderivatives by looking in lists of derivatives. See for example here: http://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_of_hyperbolic_functions

So you have $$ \int \frac{1}{\sqrt{1+x^2}}\; dx = \text{arsinh}(x) + C = \sinh^{-1}(x) + C. $$

share|improve this answer
    
Its the square root of 1+x^2. –  Lizi Dec 12 '12 at 3:28
    
Missed the sqrt in the denominator. –  Amzoti Dec 12 '12 at 3:28

The way to do it is via trigonometric substitution. If we substitute $x=\tan\theta$, then $\frac{1}{\sqrt{x^2+1}} =\frac{1}{\sqrt{1+\tan^2\theta}} = \frac{1}{\sec\theta}$ and $dx=\sec^2\theta d\theta$, so the integral becomes

$\displaystyle \int \frac{1}{\sec\theta}\sec^2\theta d\theta = \int \sec\theta d\theta$, with appropriately changed limits of integration. The indefinite integral turns out to be $\log|\sec \theta + \tan\theta|+C$.

Now using the original substitution $x=\tan\theta$ it is possible to express $\sec\theta$ and $\tan \theta$ in terms of $x$.

share|improve this answer

Your Answer

 
discard

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.