Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 '14 at 22:59
up vote 15 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.

share|cite|improve this answer
Thanks again @V-Moy. I wish I was as smart as you! :) – Dan Apr 24 '14 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 秀 Apr 24 '14 at 16:03
that's fine. thanks a lot! :) – Dan Apr 24 '14 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, $2x=3\sin t$.

share|cite|improve this answer
I win this time Prof. (>‿◠)✌ But I've to admit your method is easier than mine. Cool! +1. – Anastasiya-Romanova 秀 Apr 24 '14 at 16:09
As a general approach in a calculus course, yours is better. – André Nicolas Apr 24 '14 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 秀 Apr 24 '14 at 16:51
@V-Moy: Thank you for the correction. – André Nicolas Apr 24 '14 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 秀 Apr 24 '14 at 17:25

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

share|cite|improve this answer

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

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

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

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

share|cite|improve this answer
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 '14 at 22:52
@alexqwx I strongly believe that maths don't need words at all. – UserX Oct 5 '14 at 14:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.