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$$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$ How Can I attack this kind of problem?

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    $\begingroup$ In general remember that if you have some $\sqrt{a-b x^2}$ around typically you have to work on some substitution with $\sin t$, when instead you have $\sqrt{a+b x^2}$ try with $\sinh t$. $\endgroup$ Commented Apr 24, 2014 at 22:59

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Let $x=\cfrac{3}{2}\sin\theta$, then $dx=\cfrac{3}{2}\cos\theta\,d\theta$. \begin{align} \require{cancel} \int\frac{\sqrt{9-4x^2}}{x}\, dx&=\int\frac{\sqrt{9-4\left(\frac{3}{2}\sin\theta\right)^2}}{\cancel{\frac{3}{2}}\sin\theta}\, \cancel{\frac{3}{2}}\cos\theta\,d\theta\\ &=\int\frac{3\sqrt{1-\sin^2\theta}}{\sin\theta}\, \cos\theta\,d\theta\\ &=3\int\frac{\cos\theta}{\sin\theta}\, \cos\theta\,d\theta\\ &=3\int\frac{\cos^2\theta}{\sin\theta}\,d\theta\\ &=3\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta\\ &=3\int\frac{1}{\sin\theta}\,d\theta-3\int\sin\theta\,d\theta\\ &=3\int\frac{1}{\sin\theta}\,d\theta+3\cos\theta+C \end{align}

The last integral can be seen here, and can be done using the substitution $u = \cos \theta$ and partial fractions.

\begin{align} \int \frac{d\theta}{\sin \theta} &= \int \frac{\sin \theta}{\sin^2 \theta} d\theta\\ &= \int \frac{\sin \theta}{1 - \cos^2 \theta} d\theta\\ &= \int \frac{-du}{1 - u^2}\\ &= \int \frac{du}{u^2 - 1}\\ &= \frac{1}{2}\left(\ln\ \left|1 - u\right| - \ln\ \left|1 + u\right|\right) + C_2\\ &= \frac{1}{2}\left(\ln\ \left|1 - \cos \theta\right| - \ln\ \left|1 + \cos \theta\right|\right) + C_2 \end{align}

Hope this helps Dan.

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  • $\begingroup$ Thanks again @V-Moy. I wish I was as smart as you! :) $\endgroup$ Commented Apr 24, 2014 at 15:56
  • $\begingroup$ My pleasure @Dan. I'm not smart. ヅ BTW, I'm a bit busy here, so sorry I can't reply your comment $\endgroup$ Commented Apr 24, 2014 at 16:03
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    $\begingroup$ that's fine. thanks a lot! :) $\endgroup$ Commented Apr 24, 2014 at 16:05
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Rewrite our integral as $$\int 4x\frac{\sqrt{9-4x^2}}{4x^2}\,dx.$$ Let $9-4x^2=4u^2$. Then $x\,dx=-u\,du$, and we arrive at $$\int \frac{8u^2}{4u^2-9}\,du = \int \left( 2 + \frac{3}{2u - 3} - \frac{3}{2u + 3} \right)\ du.$$

Remark: But the answer to your question about this kind of question is probably trigonometric substitution, $2x=3\sin t$.

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    $\begingroup$ I win this time Prof. (>‿◠)✌ But I've to admit your method is easier than mine. Cool! +1. $\endgroup$ Commented Apr 24, 2014 at 16:09
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    $\begingroup$ As a general approach in a calculus course, yours is better. $\endgroup$ Commented Apr 24, 2014 at 16:13
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    $\begingroup$ Thanks for your compliment Prof. BTW, your integral should be $$ \int\frac{8u^2}{4u^2-9}\,du=\int\left(\frac{3}{2u-3}-\frac{3}{2u+3}+2\right)\,du $$ $\endgroup$ Commented Apr 24, 2014 at 16:51
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    $\begingroup$ @V-Moy: Thank you for the correction. $\endgroup$ Commented Apr 24, 2014 at 17:09
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    $\begingroup$ I really envy you Prof. If I made mistake, people here would immediately vote down my answer with no mercy but they wouldn't do that to you. In their eyes, you're like a god. \(‐^▽^‐)/ $\endgroup$ Commented Apr 24, 2014 at 17:25
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Hint: $1-\sin^2t=\cos^2t\iff9-\underbrace{9\sin^2t}_{4\,x^2}=9\cos^2t=(3\cos t)^2$

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$$(9-4x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}9^{1/2-n}(-4x^2)^{n}$$

$$\dfrac{(9-4x^2)^{1/2}}{x} = \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n}$$

$$\int \left(\dfrac{9}{x^2}\right)^{1/2-n}(-4)^{n} \;\mathrm{d}x =(-4)^{n}9^{1/2-n}\int {x}^{1-2n}\; \mathrm{d}x = (-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right) + C$$

$$\begin{align}\int \dfrac{(9-4x^2)^{1/2}}{x} \; \mathrm{d}x &= \sum_{n=0}^{\infty}\int\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n} \mathrm{d}x \\ &= \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}9^{1/2-n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{3}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}3^{2-2n}\left(\dfrac{x^{2-2n}}{2-2n}\right)\\ &= \dfrac{1}{6}\sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{(3x)^{2-2n}}{1-n}\right) + c\end{align}$$

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    $\begingroup$ Words (which connect sentences and improve articulation) are as important in maths as symbols. Please consider adding words to improve the clarity of your answers. $\endgroup$
    – beep-boop
    Commented Jul 2, 2014 at 22:52
  • $\begingroup$ @alexqwx I strongly believe that maths don't need words at all. $\endgroup$
    – UserX
    Commented Oct 5, 2014 at 14:02

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