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$\int_{2}^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$

I let $x = 2 \sec(\theta)$, then $dx = 2\sec(\theta) \tan(\theta) d \theta$

$$\int \frac{2\sec(\theta)\tan(\theta)}{2\sec(\theta)2\tan(\theta)}d\theta = \frac{1}{2}\theta + C = \frac{1}{2} \sec^{-1}\left(\frac{x}{2}\right) + C$$

Therefore the answer would be this after simplification?:

All I did is:

$$\int_{2}^{\infty} f(x)dx = \int_{2}^{3} f(x)dx + \int_{3}^{\infty} f(x)dx$$

$$\lim_{A\to2^+} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{A}^{3} + \lim_{B\to\infty} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{3}^{B}$$

$$= \frac{1}{2} (sec^{-1} (\frac{3}{2}) - 0)+ \frac{1}{2}(\pi/2 - sec^{-1}(3/2)) = \pi/4$$ Converges.

Is this right?

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  • $\begingroup$ You haven't completed your answer, so it can't be right. $\endgroup$ – DHMO Apr 16 '17 at 23:08
  • $\begingroup$ Also, you forgot to change the limits. $\endgroup$ – DHMO Apr 16 '17 at 23:08
  • $\begingroup$ Okay i simplified it. $\endgroup$ – Tinler Apr 16 '17 at 23:11
  • $\begingroup$ You still need to change your limits. $\theta$ no longer goes from $2$ to $\infty$. $\endgroup$ – DHMO Apr 16 '17 at 23:11
  • $\begingroup$ @DHMO he's changed back to $x$, surely? $\endgroup$ – Zain Patel Apr 16 '17 at 23:12
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POINT OUTS: You did not change limits. and it looks weird. But to solve via your method you just need to compute:
$$\int^\frac{\pi}{2} _0 \frac{d \theta}{2}=\frac{\pi}{4}$$
My method: let $x=\frac{1}{t}$ $$\int^0_\frac{1}{2} \frac{dt}{\sqrt{1-4t^2}}$$ Which is equivalent to $$\int^0 _\frac{1}{2} \frac{dt}{\sqrt{1-(2t)^2}}$$

Now i hope you know $$\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C$$

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  • $\begingroup$ Is my answer not correct? $\endgroup$ – Tinler Apr 16 '17 at 23:16
  • $\begingroup$ It is. @Tinler Please see the edit. $\endgroup$ – The Dead Legend Apr 16 '17 at 23:18
  • $\begingroup$ Why doesn't $u^2 - a^2$ go for $x = usec(\theta)$ ? I only know of 3 from my textbook which are sec, tan, sin. And sin is $a^2 - u^2$? $\endgroup$ – Tinler Apr 16 '17 at 23:20
  • $\begingroup$ I'm sorry, but i don't seem to get your question. What are you pointing at? @Tinler $\endgroup$ – The Dead Legend Apr 16 '17 at 23:22
  • $\begingroup$ What is wrong with my solution? I'm not saying it isn't wrong since wolfram give $\pi/4$ as well but I don't know what step I did wrong. $\endgroup$ – Tinler Apr 16 '17 at 23:23
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Hint:

If trigonometric substitution is not mandatory, choose $$\sqrt{x^2-4}=y$$

$\frac1{x\sqrt{x^2-4}}=\frac x{x^2\sqrt{x^2-4}}$

Now change the limits accordingly.

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hint the integral is convergent since

$$f (x)\sim \frac {1}{x^2}\;\;(x\to\infty) $$

$$f(x)\sim \frac{1}{4\sqrt{x-2}}\;\;(x\to2^+)$$

to find its value, put $$x=2\cosh (t) =2\frac{e^t+e^{-t}}{2}.$$ $x=2\implies t=0$

$x\to \infty\implies t \to \infty $.

on the other hand,

$$dx=2\sinh (t) $$ and $$\cosh^2 (t)-\sinh^2 (t)=1.$$

thus

$$x\sqrt {x^2-4}=4\cosh(t)\sinh(t). $$

the integral becomes

$$\int_0^\infty \frac{dt}{e^t+e^{-t}} $$ $$\color {green}{\boxed {=[\arctan(e^t)]_0^\infty=\frac {\pi}{4}}}$$

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