1
$\begingroup$

$\newcommand{\arcsec}{\operatorname{arcsec}}$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.

$\endgroup$
2
  • $\begingroup$ $d\theta = \frac{3}{\sqrt{x^2-9} |x|} dx \iff dx= \frac{\sqrt{x^2-9} |x|}3d\theta$ $\endgroup$ Dec 27, 2020 at 17:59
  • $\begingroup$ See my edits to this question for proper MathJax usage. $\endgroup$ Dec 27, 2020 at 18:35

6 Answers 6

4
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ yes, that's what I did but it should work the same with trig subsitution, no? $\endgroup$
    – John Liu
    Dec 27, 2020 at 18:53
  • $\begingroup$ I have added a 3rd approach... does that help you to reconcile? $\endgroup$
    – Doug M
    Dec 27, 2020 at 19:04
  • $\begingroup$ sorry i didnt fully read it, my bad. yeah, your solution helps a lot $\endgroup$
    – John Liu
    Dec 28, 2020 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, 2020 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, 2021 at 16:51
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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, 2020 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, 2020 at 19:35
1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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, 2020 at 18:38
  • $\begingroup$ Maybe this helps you en.wikipedia.org/wiki/Hyperbolic_functions $\endgroup$
    – user0102
    Dec 27, 2020 at 18:39
0
$\begingroup$

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: $$I=3^5\int\cosh^3v\sqrt{\cosh^2v-1}\sinh v dv$$ $$I=3^5\int\cosh^3v\sinh^2vdv$$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Yes, I did u-substitution and integration by parts, but I was looking for a trig sub solution. $\endgroup$
    – John Liu
    Dec 28, 2020 at 17:34
0
$\begingroup$

\begin{aligned} & \int x^{3} \sqrt{x^{2}-9} d x \\ =& \int \frac{x^{3}\left(x^{2}-9\right)}{\sqrt{x^{2}-9}} d x \\ =& \int x^{2}\left(x^{2}-9\right) d\left(\sqrt{x^{2}-9}\right) \\ =& \int\left(y^{2}+9\right) y^{2} d y,\text { where } y=\sqrt{x^{2}-9} \\ =& \frac{y^{5}}{5}+3 y^{3}+c \\ =& \frac{y^{3}}{5} \cdot\left(y^{2}+15\right)+c \\ =& \frac{\left(x^{2}-9\right)^{\frac{3}{2}}}{5}\left(x^{2}+6\right)+c \end{aligned}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.