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I would like to find

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

I have found that it is equivalent to $$ \int \frac{\sqrt{1 + x^2}}{x}dx $$ but I am not sure what to do about it. With trig substitution $x = \tan(x)$ I get $$ \int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta $$ but that seems to be a dead end.

6 Answers 6

4

$$ \int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta $$ $$ =\int \frac{\sin(\theta)}{\cos^2(\theta)(1-\cos^2(\theta))}d\theta $$ then use $u=\cos(\theta)$ and partial fractions on the result.

4

Let $x^2+1=u^2$, so that $2x\,dx=2u\,du$, which implies

$${dx\over x}={u\,du\over x^2}={u\over u^2-1}du$$

Then

$$\int{\sqrt{1+x^2}\over x}dx=\int{u^2\over u^2-1}du={1\over2}\int\left(2+{1\over u-1}-{1\over u+1} \right)du=u+{1\over2}\ln\left(u-1\over u+1 \right)+C\\=\sqrt{x^2+1}+{1\over2}\ln\left(\sqrt{x^2+1}-1\over\sqrt{x^2+1}+1 \right)+C$$

2

and $$ \int \frac{1}{\sin(\theta)\cos^2(\theta)}d\theta= \int \frac{\sin\theta}{\sin^2(\theta)\cos^2(\theta)}d\theta= \int \frac{-d(\cos\theta)}{(1-\cos^2\theta)\cos^2(\theta)}d\theta=... $$

2

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

With substitution $u=x^{2}$, $du=2xdx$.

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

With substitution $v=\sqrt{u+1}$, $dv=\frac{1}{2v}du$.

\begin{align} \frac{1}{2} \int \frac{u+1}{u}du &=\frac{1}{2} \int \frac{2v^{2}}{v^{2}-1}dv\\[6px] &=v-\operatorname{arctanh} (v)+C\\[6px] &=\sqrt{x^{2}+1}-\operatorname{arctanh}(\sqrt{x^{2}+1})+C. \end{align}

  • 0
    Upon your first suggested substitution, $$dx= \frac{du}{2x} = \frac{du}{2\sqrt u} $$ $$\int \frac{\sqrt{1 + x^2}}{x}dx=\int \frac{\sqrt{u+1}}{2\sqrt u}du$$ – 2017-02-06
2

I get $\int \frac{sec \theta}{tan \theta}d \theta$ from your simplification

Which gives $\int \frac{1/cos \theta}{sin \theta / cos \theta}d \theta$

= $\int \frac{1}{sin \theta}d \theta$

= $\int csc \theta d \theta$

which integrates similarly to $sec\ \theta$

and then the fun of back-substituting.

  • 0
    I am not sure I see how you get the $\sec(\theta) / \tan(\theta)$ – 2017-02-07
  • 0
    You had $$\int \frac{\sqrt{ 1 + x^2}}{x} dx$$ – 2017-02-07
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    Yes, but don't you also need to do $dx = ... d\theta$? – 2017-02-07
  • 1
    Sorry, keep hitting return too soon. I think I had a typo there and intended to correct, sorry. My error. Still getting used to this system. You substituted $x = tan(\theta)$ so $dx = sec^2 (\theta) d \theta$ $ \sqrt{1 + x^2} = \sqrt{1 + tan^2 (\theta)} = \sqrt{sec^2(\theta} } = sec(\theta)$ You now have $$ \int \frac{ sec \theta * sec^2 \theta} {tan \theta} d \theta $$ so yes I slipped up somewhere. Had another approach to this and lost it, The sinh sub suggested in another answer is a good one. – 2017-02-07
2

Set $x^{-1}=\sinh t$, so $\sqrt{1+x^{-2}}=\cosh t$. Then $$ dx=-\frac{\cosh t}{\sinh^2t}\,dt $$ and the integral becomes $$ -\int\frac{\cosh^2t}{\sinh^2t}\,dt= -\int\frac{1+\sinh^2t}{\sinh^2t}\,dt=\frac{\cosh t}{\sinh t}-t+c= \sqrt{1+x^2}-\operatorname{arsinh}\frac{1}{x}+c $$ You can find a more explicit expression for $\operatorname{arsinh}\frac{1}{x}$ by setting $$ \frac{1}{x}=\frac{e^t-e^{-t}}{2} $$ or $$ xe^{2t}-2e^t-x=0 $$ so $$ e^t=\frac{1+\sqrt{1+x^2}}{x} $$ The final antiderivative is $$ \sqrt{1+x^2}-\log\frac{1+\sqrt{1+x^2}}{x}+c $$

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    This is very nice. I like the use of the hyperbolic functions ending with something with a log. Thanks! – 2017-02-07