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Can anyone show how to calculate $$\int\sqrt{1+u^2}\,du?$$ I can't calculate it.

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  • $\begingroup$ Calculate would be a better word. Sorry for confusing you. $\endgroup$ – CALC-FATH Dec 3 '13 at 2:57
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    $\begingroup$ Stewart tells us to do a trig sub but $\sec^3\theta$ leaves a bit of work...Kaster's approach is much more efficient for whoever actually tries to carry out the computation til the end. $\endgroup$ – Julien Dec 3 '13 at 3:10
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    $\begingroup$ I don't normally think of this integral as anything special, but having glanced at the responses below I'm filing this one away as a future teaching tool, since it so wonderfully demonstrates the versatility of integration tactics. $\endgroup$ – David H Dec 3 '13 at 3:45
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Setting $$ u=\sinh x:=\frac{e^x-e^{-x}}{2}, $$ we have \begin{eqnarray} \int\sqrt{1+u^2}\,du&=&\int\cosh x\sqrt{1+\sinh^2x}\,dx=\int\cosh^2x\,dx=\frac12\int(1+\cosh2x)\,dx\\ &=&\frac{x}{2}+\frac14\sinh2x+C=\frac{x}{2}+\frac12\sinh x\cosh x+C\\ &=&\frac12\ln(u+\sqrt{1+u^2})+\frac12u\sqrt{1+u^2}+C. \end{eqnarray}

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  • $\begingroup$ While the tan substitution is the most obious one, I really like this one too. Because it is more (or should I say different) work. Versatility counts too. So upvote! $\endgroup$ – imranfat Dec 3 '13 at 3:10
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Let $u = \sinh x$, then \begin{align} \int \sqrt{1+u^2}du &= \int \cosh x \cdot \cosh x dx = \int \cosh^2x dx = \int \frac {1+\cosh 2x}2 dx = \\ &= \frac x2 + \frac {\sinh 2x}4 + C = \frac {\text{arcsinh } u}2 + \frac {u \sqrt{1+u^2}}2 + C \end{align}

PS

To verify: WA

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Let $\displaystyle I = \int \sqrt{1+u^2}du = \int \sqrt{1+u^2}\cdot 1\;du$

Using Integration by parts

$\displaystyle I = \sqrt{1+u^2}\cdot u-\int \frac{u}{\sqrt{1+u^2}}\cdot udu = \sqrt{1+u^2}\cdot u-\int\frac{(1+u^2)-1}{\sqrt{1+u^2}}du$

$\displaystyle I = \sqrt{1+u^2}\cdot u-I+\int\frac{1}{\sqrt{1+u^2}}du$

$\displaystyle 2I = \sqrt{1+u^2}\cdot u+J$

where $\displaystyle J = \int \frac{1}{\sqrt{1+u^2}}du$

for Calculation of $J$

Let $1+u^2 = v^2$ and $\displaystyle udu = vdv\Rightarrow \frac{du}{v} = \frac{dv}{u} = \frac{d(u+v)}{(u+v)}$

(above Using ratio and Proportion )

So $\displaystyle J = \int\frac{du}{v} = \int\frac{d(u+v)}{(u+v)} = \ln \left|u+v\right|+C$

So $\displaystyle J = \ln \left|u+\sqrt{1+u^2}\right|+C$

So $\displaystyle I = \frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln \left|u+\sqrt{1+u^2}\right|+D$

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To integrate this we would use Trigonometric substitution. Recall your substitution rules for this. In this case, we would let $u = 1\tan(\theta)$ and then continue to find $d\theta$ so we may do the full substitution.

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  • $\begingroup$ It'd be worth noting how you can figure out these substitution relationships by using the $sin^2 + cos^2 = 1$ identity. Then you don't have to memorize anything ;) $\endgroup$ – GraphicsMuncher Dec 3 '13 at 3:01
  • $\begingroup$ I really would like to see how you'll proceed after that! $\endgroup$ – Mercy King Dec 3 '13 at 3:16
  • $\begingroup$ @Mercy It's kinda like your hyperbolic substitution, but ends up with the integral of $\sec^3x$. If you've either a) memorized the integral of $\sec^3x$, or b) become used to computing integrals like that, so it doesn't phase you, then you're all set. :) $\endgroup$ – apnorton Dec 3 '13 at 3:33
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Let $ u = \tan \theta $. Then, $ \mathrm{d}u = \sec^2 \theta \, \mathrm{d}\theta $. Also, note that $$ \sqrt {1 + u^2} = \sqrt {1 + \tan^2 \theta} = \sec \theta. $$Use this to finish.

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    $\begingroup$ I really would like to see how you'll proceed after that! $\endgroup$ – Mercy King Dec 3 '13 at 3:16
  • $\begingroup$ @Mercy I simply did not show my work because the OP hasn't shown any him/herself. But it becomes simple Integration by Parts after that. $\endgroup$ – Ahaan S. Rungta Dec 3 '13 at 3:18
  • $\begingroup$ Are you sure about that? $\endgroup$ – Mercy King Dec 3 '13 at 3:20
  • $\begingroup$ @Mercy Yes. I have tried it. $\endgroup$ – Ahaan S. Rungta Dec 3 '13 at 3:34

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