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Compute $$I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$$

The way I approach this problem was to:

  1. Set $u=\sqrt{x-2}$ and arrive at $$I=2\int\frac{u^2(u^2+1)}{\sqrt{u^2-4}}\mathrm du$$
  2. Set $u=2\sec t\implies\mathrm du=2\sec t\tan t\mathrm dt$ to get $I=8\int\sec^3(t)(4\sec^2(t)+1)\mathrm dt$

Now this integral involves $\sec^5t$ and $\sec^3t$ which does not make me very happy. I think there should be a simpler method. Can anyone show me simpler steps? (the first few steps/substitutions would suffice)

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Write $x+3=(x-2)+5$ and use $\int\sqrt{x^2-a^2}dx$. Now,

$$\begin{align} \\ \sqrt{\dfrac{x+2}{x-2}} &= \dfrac{x+2}{\sqrt{x^2-4}} \\ &=\dfrac x{\sqrt{x^2-4}}+\dfrac2{\sqrt{x^2-2^2}} \end{align}$$

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  • $\begingroup$ Got it. Thanks! (will accept this answer in 2-3 days to allow for other people to attempt) $\endgroup$ – Gaurang Tandon Mar 10 '18 at 4:05
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When you end up with $\int\sec\theta\,d\theta$ it often means that you would have been better off making a $\cosh\theta$ substitution instead. Here we are going to use $x=2\cosh\theta$ so $\sqrt{x^2-4}=2\sinh\theta$ and $dx=2\sinh\theta\,d\theta$ $$\begin{align}\int(x+3)\sqrt{\frac{x+2}{x-2}}dx&=\int\frac{(x+3)(x+2)}{\sqrt{x^2-4}}dx\\ &=\int\frac{(2\cosh\theta+3)(2\cosh\theta+2)}{2\sinh\theta}2\sinh\theta\,d\theta\\ &=\int\left(4\cosh^2\theta+10\cosh\theta+6\right)d\theta\\ &=\int(2\cosh2\theta+10\cosh\theta+8)d\theta\\ &=\sinh2\theta+10\sinh\theta+8\theta+C_1\\ &=2\sinh\theta\cosh\theta+10\sinh\theta+8\theta+C_1\\ &=2\frac{\sqrt{x^4-4}}2\frac x2+10\frac{\sqrt{x^2-4}}2+8\cosh^{-1}\left(\frac x2\right)+C_1\\ &=\left(\frac12x+5\right)\sqrt{x^2-4}+8\ln\left(\frac x2+\sqrt{\frac{x^2}4-1}\right)+C_1\\ &=\left(\frac12x+5\right)\sqrt{x^2-4}+8\ln\left(x+\sqrt{x^2-4}\right)+C\end{align}$$ Where $C=C_1-\ln2$. Differentiation verifies this result.


Given that the OP doesn't know about the hyperbolic functions: $$\begin{align}\cosh x&=\frac{e^x+e^{-x}}2\\ \sinh x&=\frac{e^x-e^{-x}}2\end{align}$$ So $$\begin{align}\cosh^2x-\sinh^2x&=\frac{e^{2x}+2+e^{-2x}}4-\frac{e^{2x}-2+e^{-2x}}4=1\\ \cosh^2x+\sinh^2x&=\frac{e^{2x}+2+e^{-2x}}4+\frac{e^{2x}-2+e^{-2x}}4=\frac{e^{2x}+e^{-2x}}2=\cosh2x\end{align}$$ On addition of the last $2$ formulas we have $$2\cosh^2x=\cosh2x+1$$ And finally there is $$\begin{align}\frac d{dx}\cosh x&=\frac d{dx}\frac{e^x+e^{-x}}2=\frac{e^x-e^{-x}}2=\sinh x\\ \frac d{dx}\sinh x&=\frac d{dx}\frac{e^x-e^{-x}}2=\frac{e^x+e^{-x}}2=\cosh x\end{align}$$ Oh, there is one more thing: if $y=\cosh^{-1}x$ then $$x=\cosh y=\frac{e^y+e^{-y}}2$$ So $$e^{2y}-2xe^y+1=0$$ Solving for $y$, $$y=\cosh^{-1}x=\ln\left(x+\sqrt{x^2-1}\right)$$ Now you know!

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  • $\begingroup$ Thanks! But I've not yet been taught the hyperbolic functions route. But, this answer will definitely help others those who know it. So, thanks! $\endgroup$ – Gaurang Tandon Mar 10 '18 at 4:48
  • $\begingroup$ I appended a brief lecture on hyperbolic functions that contains all the identities necessary for this problem. $\endgroup$ – user5713492 Mar 10 '18 at 5:32
  • $\begingroup$ Thanks! I understood everything. That was very helpful :D $\endgroup$ – Gaurang Tandon Mar 10 '18 at 6:03
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Note: This is a long, long stretch.

Final Answer:$$\dfrac{16\ln\left(\left|\sqrt{\frac{x^2}4-1}\right|\right)+(x+10)\sqrt{x^2-4}}2+C$$

After applying long division on $\dfrac{(x+3)\sqrt{x^2-4}}{x-2}$ and rewriting the integral, you get: $$5\int\frac{\sqrt{x^2-4}}{x-2}dx+\int\sqrt{x^2-4} dx$$

Let $x=2\sec(u)\rightarrow u=\text{arcsec}\left(\dfrac x2\right) \rightarrow dx=2\sec(u)\tan(u)du$.$$\int\frac{\sqrt{x^2-4}}{x-2}dx=\int\dfrac{2\sec(u)\sqrt{4\sec^2(u)-4}\tan(u)}{2\sec(u)-2}=2\left(\int\cos(u)\sec^2(u)du+\int\sec^2(u)du \right)$$

$$=2\left(\int\sec(u)+\int\sec^2(u)du \right)=2\Big{(}\ln\big{(}\tan(u)+\sec(u)\big{)}+\tan(u)\Big{)}$$

Substituting from $u$ to $x$: $$2\Big{(}\ln\big{(}\tan(u)+\sec(u)\big{)}+\tan(u)\Big{)}=2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)+2\sqrt{\frac {x^2}4-1}$$ Then: $$\int\sqrt{x^2-4}=x\sqrt{\frac {x^2}4-1}-2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)$$

So putting together our steps:

$$5\int\frac{\sqrt{x^2-4}}{x-2}dx+\int\sqrt{x^2-4} dx=$$$$5\cdot \left(2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)+2\sqrt{\frac {x^2}4-1}\right)+x\sqrt{\frac {x^2}4-1}-2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)$$

Which finally helps you arrive at the answer. Whew!

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  • $\begingroup$ That substitution seems to be exactly what I was trying to do as well. Still, thanks for illustrating how unwieldy it actually is. $\endgroup$ – Gaurang Tandon Mar 10 '18 at 6:05
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I think the easiest substitution should be $x= 2\cos2 \alpha$ and then ,you get

$\int i(2\cos2 \alpha +3)\cot \alpha \mathrm{d}(\ 2cos2 \alpha)$

$= \int i(2\cos2 \alpha +3)\cot \alpha (-4 \sin2\alpha)\mathrm{d}\alpha $

Now you can proceed, with trigonometric simplifications.

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  • $\begingroup$ This will save a lot of time. $\endgroup$ – Vishaal Selvaraj Mar 10 '18 at 5:18
  • $\begingroup$ Hey thanks. Did you mean $\cos^2\alpha$ or $\cos(2\alpha)$? $\endgroup$ – Gaurang Tandon Mar 10 '18 at 5:22
  • $\begingroup$ @GaurangTandon $cos2 \alpha$ because $\frac {1-cos2x}{1+cos2x} = tan^{2}x$ $\endgroup$ – Vishaal Selvaraj Mar 10 '18 at 5:27
  • $\begingroup$ Yes, but you end up with $$\frac{\cos2x+1}{\cos2x-1}<0$$ $\endgroup$ – user5713492 Mar 10 '18 at 5:29
  • $\begingroup$ Now, I don't get this. Where'd you get the $i$ from? $\endgroup$ – Gaurang Tandon Mar 10 '18 at 6:04

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