# Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$

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

• 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 $\sinh t$. Commented Apr 24, 2014 at 22:59

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.

• Thanks again @V-Moy. I wish I was as smart as you! :) Commented Apr 24, 2014 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 Commented Apr 24, 2014 at 16:03
• that's fine. thanks a lot! :) Commented Apr 24, 2014 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, $2x=3\sin t$.

• I win this time Prof. (>‿◠)✌ But I've to admit your method is easier than mine. Cool! +1. Commented Apr 24, 2014 at 16:09
• As a general approach in a calculus course, yours is better. Commented Apr 24, 2014 at 16:13
• Thanks for your compliment Prof. BTW, your integral should be $$\int\frac{8u^2}{4u^2-9}\,du=\int\left(\frac{3}{2u-3}-\frac{3}{2u+3}+2\right)\,du$$ Commented Apr 24, 2014 at 16:51
• @V-Moy: Thank you for the correction. Commented Apr 24, 2014 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. \（‐＾▽＾‐）/ Commented Apr 24, 2014 at 17:25

Hint: $1-\sin^2t=\cos^2t\iff9-\underbrace{9\sin^2t}_{4\,x^2}=9\cos^2t=(3\cos t)^2$

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

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

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

\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}

• 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. Commented Jul 2, 2014 at 22:52
• @alexqwx I strongly believe that maths don't need words at all. Commented Oct 5, 2014 at 14:02