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$\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.

Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$ I then substituted $x = 3\sec\theta$, $\theta = \arcsec(\frac{x}{3})$, and $d\theta = \frac{3}{\sqrt{x^2-9} |x|} \, dx$. Plugging these values into the original integral, $$\int 27\sec^3\theta\sqrt{9\sec^2\theta - 9}\frac{3}{\sqrt{9\sec^2\theta-9} |3\sec\theta|} \, d\theta$$ $$\int \frac{81\sec^3\theta3\tan\theta}{3\tan\theta |3\sec\theta|} \, d\theta$$ $$\int \frac{81\sec^3\theta}{|3\sec\theta|} \, d\theta$$ $$\int 27\sec^2\theta \, d\theta$$ $$27\tan\theta + C$$ Now substituting $x$ back in and simplifying: $$27\tan(\arcsec(\frac{x}{3})) + C$$ $$9\operatorname{sgn}(x)\sqrt{x^2-9} + C$$

This does not seem close at all to the solution I found by $u$-substitution, $$\frac{(x^2-9)^\frac{3}{2}(x^2+6)}{5} + C$$

I am relatively new to integration so I think I made a mistake. What did I do wrong? Any help appreciated.

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  • $\begingroup$ $d\theta = \frac{3}{\sqrt{x^2-9} |x|} dx \iff dx= \frac{\sqrt{x^2-9} |x|}3d\theta$ $\endgroup$ – J. W. Tanner Dec 27 '20 at 17:59
  • $\begingroup$ See my edits to this question for proper MathJax usage. $\endgroup$ – Michael Hardy Dec 27 '20 at 18:35
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With a u-substitution:
$\int x^3\sqrt{x^2-9} \ dx\\ \int \frac 12 x^2\sqrt{x^2-9} (2x\ dx)\\ u = x^2 - 9\\ x^2 = u+9\\ du = 2x\ dx\\ \frac 12\int (u+9)\sqrt{u}\ du\\ \frac 12(\frac 23 (9u^\frac 32) + \frac 25 u^\frac 52)+ C\\ 3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$

With a trig substitution:
$\int x^3\sqrt{x^2-9} \ dx\\ x = 3\sec\theta\\ dx = 3\sec\theta\tan\theta\\ 3^5\int \sec^4\theta\tan^2\theta\ d\theta\\ 3^5\int (\tan^2\theta\ + \tan^4\theta)\sec^2\theta d\theta\\ 3^5 (\frac 13\tan^3\theta\ + \frac 15\tan^5\theta) + C\\ 3^5 (\frac 13 (\frac {x^2}{9}-1)^\frac 32 + \frac 15(\frac {x^2}{9}-1)^\frac 52) + C\\ 3^5 (\frac 13 \frac {(x^2-9)^\frac 32}{3^3} + \frac 15\frac {(x^2-9)^\frac 52}{3^5}) + C\\ 3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$

If you want to use the substitution

$\theta = \sec^{-1} \frac {x}{3}\\ dx = \frac {3}{x\sqrt{x^2-9}}$

$\int x^3\sqrt{x^2-9} \ dx\\ \int \frac 13 x^4(x^2-9)\left(\frac {3}{x\sqrt{x^2-9}}\ dx\right)\\ \frac 13 \int (3^4\sec^4 \theta) (3^2\tan^2\theta)\ d\theta\\ $

And continue as above.

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  • $\begingroup$ yes, that's what I did but it should work the same with trig subsitution, no? $\endgroup$ – John Liu Dec 27 '20 at 18:53
  • $\begingroup$ I have added a 3rd approach... does that help you to reconcile? $\endgroup$ – Doug M Dec 27 '20 at 19:04
  • $\begingroup$ sorry i didnt fully read it, my bad. yeah, your solution helps a lot $\endgroup$ – John Liu Dec 28 '20 at 3:54
  • $\begingroup$ Actually one more question, how do $tan^3(arcsec(x/3))$ and $tan^5(arcsec(x/3))$ turn into those values? $\endgroup$ – John Liu Dec 28 '20 at 4:13
  • $\begingroup$ @DougM excellent explanation, could you give me a bit help to this similar problem if possible :) math.stackexchange.com/questions/4081654/… $\endgroup$ – John Apr 4 at 16:51
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The mistake is in the differential $d\theta$. You accidentally switched $d\theta$ for $dx$. If you fix that it should be OK. Alternatively, differentiate $x=3\sec{\theta}$ on both sides, so you obtain: $$dx=3\sec{\theta}\tan{\theta}\,d\theta$$

Try to do the rest from here

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  • $\begingroup$ I tried this and got $\frac{243}{4}sec^3(\frac{x}{3}) + C$. Is this correct? It doesn't look much like the other answer either. $\endgroup$ – John Liu Dec 27 '20 at 18:52
  • $\begingroup$ @JohnLiu If you follow the steps in Doug M's answer it should get you to: $$\int x^3\sqrt{x^2-9} \ dx=3(x^2-9)^\frac 32 + \frac 15(x^2-9)^\frac 52 + C$$ $\endgroup$ – Clerni Dec 27 '20 at 19:35
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HINT

Make the change of variable $x = 3\cosh(u)$. Thus we get that \begin{align*} \int x^{3}\sqrt{x^{2} - 9}\mathrm{d}x & = 243\int\cosh^{3}(u)\sinh(u)\sqrt{\cosh^{2}(u) - 1}\mathrm{d}u\\\\ & = 243\int\cosh^{3}(u)\sinh^{2}(u)\mathrm{d}u\\\\ & = 243\int\cosh(u)\cosh^{2}(u)\sinh^{2}(u)\mathrm{d}u\\\\ & = 243\int\cosh(u)(1 + \sinh^{2}(u))\sinh^{2}(u)\mathrm{d}u\\\\ & = 243\int[\cosh(u)\sinh^{2}(u) + \cosh(u)\sinh^{4}(u)]\mathrm{d}u\\\\ & = 81\sinh^{3}(u) + \frac{243\sinh^{5}(u)}{5} + C \end{align*}

Can you take it from here?

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  • $\begingroup$ not really, i don't know much about hyperbolic functions, integrating them, or their identities. Is there another way? Although this does seem simpler $\endgroup$ – John Liu Dec 27 '20 at 18:38
  • $\begingroup$ Maybe this helps you en.wikipedia.org/wiki/Hyperbolic_functions $\endgroup$ – APCorreia Dec 27 '20 at 18:39
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sometimes its easier to break it up into other expressions first: $$I=\int x^3\sqrt{x^2-9}$$ $x=3u\Rightarrow dx=3du$ $$I=\int(3u)^3\sqrt{3^2(u^2-1)}(3du)$$ $$I=3^5\int u^3\sqrt{u^2-1}du$$ now we want to look for an identity that matches this, we know:

  1. $\cosh^2\theta-\sinh^2\theta=1$
  2. $\cos^2\theta+\sin^2\theta=1$

so we want to choose one of these that can be rearranged to our form, the nicest would be: $\cosh^2\theta-1=\sinh^2\theta$ so:

$u=\cosh(v)\Rightarrow du=\sinh(v)dv$ now replace everything: $$I=3^5\int\cosh^3v\sqrt{\cosh^2v-1}\sinh v dv$$ $$I=3^5\int\cosh^3v\sinh^2vdv$$

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Wouldn't be integration by parts much easier? Start with $$\frac12\int x^2\cdot2x\sqrt{x^2-9}\,dx.$$

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  • $\begingroup$ Yes, I did u-substitution and integration by parts, but I was looking for a trig sub solution. $\endgroup$ – John Liu Dec 28 '20 at 17:34

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