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

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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 $\sinh t$. –  Matteo Italia Apr 24 at 22:59

4 Answers 4

up vote 13 down vote accepted

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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Thanks again @V-Moy. I wish I was as smart as you! :) –  Dan Apr 24 at 15:56
My pleasure @Dan. I'm not smart. ヅ BTW, I'm a bit busy here, so sorry I can't reply your comment –  Anastasiya-Romanova Apr 24 at 16:03
that's fine. thanks a lot! :) –  Dan Apr 24 at 16:05

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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I win this time Prof. (>‿◠)✌ But I've to admit your method is easier than mine. Cool! +1. –  Anastasiya-Romanova Apr 24 at 16:09
As a general approach in a calculus course, yours is better. –  André Nicolas Apr 24 at 16:13
Thanks for your compliment Prof. BTW, your integral should be $$ \int\frac{8u^2}{4u^2-9}\,du=\int\left(\frac{3}{2u-3}-\frac{3}{2u+3}+2\right)\,du $$ –  Anastasiya-Romanova Apr 24 at 16:51
@V-Moy: Thank you for the correction. –  André Nicolas Apr 24 at 17:09
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. \(‐^▽^‐)/ –  Anastasiya-Romanova Apr 24 at 17:25

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

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

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

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

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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. –  alexqwx Jul 2 at 22:52

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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