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indefinite integral $$\int\sqrt {x^2 + a^2} dx$$

After some transformations and different substitution, I got stuck at this

$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$

I am not sure I am getting the first step correct. Tried substituting x as a times tan theta but that doesn't help either.

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  • $\begingroup$ Please avoid displaystyle math in titles, see here for meta thread on this issue, in this case I have edited the question to be inline. $\endgroup$ – Alice Ryhl Sep 17 '14 at 20:38
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Consider the integral \begin{align} I = \int \sqrt{x^{2}+ a^{2}} \, dx. \end{align} Make the substitution $x = a \sinh(t)$, $dx = a \cosh(t) dt$, it is seen that \begin{align} I &= a \int \sqrt{ a^{2} (1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\ &= a^{2} \int \sqrt{\cosh^{2}(t)} \cosh(t) \, dt \\ &= a^{2} \int \cosh^{2}(t) \, dt \\ &= \frac{a^{2}}{2} \int (1 + \cosh(2t)) dt \\ &= \frac{a^{2}}{2} \left[ t + \frac{1}{2} \sinh(2t) \right] \\ &= \frac{a^{2}}{2} \left[ t + \sinh(t) \cosh(t) \right]. \end{align} Now back substitute to obtain \begin{align} \int \sqrt{x^{2}+ a^{2}} \, dx &= \frac{a^{2}}{2} \left[ \sinh^{-1}(x/a) + (x/a) \cosh(\sinh^{-1}(x/a)) \right] \\ &= \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \sinh^{-1}\left( \frac{x}{a} \right). \end{align}

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  • $\begingroup$ Ah,I should probably learn to use more hyperbolic functions. This is the correct answer, wolfram though gives the second part as a logarithmic function instead of inverse sinh. But I know both are correct. Thanks. How do I mark this as the accepted answer (new to posting here) $\endgroup$ – Urmish Shah Sep 17 '14 at 21:33
  • $\begingroup$ @Urmish There is a check mark symbol under the up and down arrows. If you click on the check mark it will make it the question proposer's best chosen solution. As to the rest, this is what is great about this site. Sometimes it is just a missing method that solves the problems. Once seen it just makes sense. $\endgroup$ – Leucippus Sep 17 '14 at 21:45
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(I assume $a>0$, which is not restrictive.)

This can be treated in a way very similar to $\int\sqrt{a^2-x^2}\,dx$. Set $x=a\sinh t$, so $dx=a\cosh t\,dt$ and $$ \sqrt{a^2\sinh^2t+a^2}=a\sqrt{\sinh^2t+1}=a\cosh t $$ Thus you have to compute $$ \int\cosh^2t\,dt $$ The fundamental relation is $$ \cosh^2t-\sinh^2t=1 $$ so $$ \int\cosh^2t\,dt-\int\sinh^2t\,dt=t\rlap{\qquad(*)} $$ (I'll omit the constant of integration). But, integrating by parts, $$ \int\sinh t\sinh t\,dt=\cosh t\sinh t-\int\cosh t\cosh t\,dt $$ or $$ \int\sinh^2 t\,dt=\cosh t\sinh t-\int\cosh^2t\,dt $$ and, replacing in $(*)$ we get $$ 2\int\cosh^2t\,dt=t-\cosh t\sinh t $$ Now it's just a problem of back substitution: $\sinh t=\frac{x}{a}$ and $$ \cosh t=\frac{\sqrt{x^2+a^2}}{a} $$ while, from $$ a\frac{e^t-e^{-t}}{2}=x $$ we get $$ ae^2t-2xe^t-a=0 $$ or $$ e^t=\frac{x+\sqrt{x^2+a^2}}{a} $$ and so $$ t=\log\bigl(x+\sqrt{x^2+a^2}\,\bigr)-\log a $$

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Let $x=a\tan\theta$, so $dx=a\sec^{2}\theta d\theta$ to get

$\int\sqrt{x^2+a^2}\;dx=a^2\int\sec^{3}\theta\;d\theta$. Using integration by parts with $u=\sec\theta$ and $dv=\sec^{2}\theta\;d\theta$ gives

$\hspace{.6 in}\sec^{3}\theta\;d\theta=\frac{1}{2}\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C$,

so $\int\sqrt{x^2+a^2}=\frac{a^2}{2}\left(\frac{\sqrt{x^2+a^2}}{a}\frac{x}{a}+\ln\left|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}\right|\right)+C$

$\hspace{.9 in}=\frac{1}{2}\left({x\sqrt{x^2+a^2}}+a^2\ln\left(\sqrt{x^2+a^2}+x\right)\right)+C$

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  • $\begingroup$ I want to accept this answer and the other one where I left a comment (uses hyperbolic trig functions) If any mods are looking at this, these the the correct answers among some others and a few wrong ones as well. I don't have the power to vote since I am new user. $\endgroup$ – Urmish Shah Sep 17 '14 at 22:12
  • $\begingroup$ Leucippus gave his/her answer before mine, and it is in some ways simpler, so you should be able to click on the check mark below the arrows to accept that answer. $\endgroup$ – user84413 Sep 17 '14 at 22:17
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Here we have another way to see this: $$ \int \sqrt{x^2+a^2} dx $$ using the substitution $$ t=x+\sqrt{x^2+a^2}\\ \sqrt{x^2+a^2}=t-x $$ and squaring we have $$ a^2 =t^2-2tx\\ x=\frac{t^2-a^2}{2t}. $$ Finally we can use: $$ dx=\frac{2t(t)-(t^2-a^2)(1)}{2t^2}dt = \frac{t^2+a^2}{2t^2}dt\\ \sqrt{x^2+a^2}=t-\frac{t^2-a^2}{2t}=\frac{t^2+a^2}{2t}. $$ Thus: $$ \int \sqrt{x^2+a^2} dx = \int \frac{t^2+a^2}{2t} \frac{t^2+a^2}{2t^2}dt=\int \frac{(t^2+a^2)^2}{4t^3}dt $$ which is elementary, if we expand the square of the binomial: $$ \int\frac{t}{4}dt+\int\frac{a^2}{2t}dt+\int\frac{a^4}{4t^3}dt=\frac{t^2}{8}+a^2\ln\sqrt{t}-\frac{a^4}{16t^4}, $$ where as stated $t=x+\sqrt{x^2+a^2}.$

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$$\sqrt{x^2+a^2}dx = a^2\sqrt{\Big(\frac{x}{a}\Big)^2+1}\,d\frac{x}{a}.$$ Set $\sinh\theta=\frac{x}{a}$. $$\sqrt{1+\sinh^2\theta}d\sinh\theta=\cosh^2\theta d\theta=\frac{1}{2}(\cosh2\theta+1)d\theta=\frac{1}{4}d(\sinh2\theta+2\theta).$$

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