$$\int \frac{\sqrt{9-4x^2}}{x}dx$$

I know I can write it as $$\int \frac{\sqrt{3^2-2^2x^2}}{x}dx$$

And use $x=\frac{3}{2}sin\theta\rightarrow2x=3sin\theta d\theta $ And $2dx=3cos\theta d\theta\rightarrow dx=\frac{3}{2}cos\theta d\theta$

So I get $$\int \frac{\sqrt{9-9sin^2\theta}}{\frac{3}{2}sin\theta}\frac{3}{2}cos\theta d\theta=\int \frac{3\sqrt{1-sin^2\theta}}{sin\theta}\cos\theta d\theta=\int \frac{3*cos\theta}{sin\theta}\cos\theta d\theta=3\int \frac{cos^2\theta}{sin\theta}d\theta$$

How should I continue?

Can I use $u=cos\theta$ and $du=-sin\theta$ so it is $ -\int \frac{u^2}{du}$?

  • 1
    $\begingroup$ Re: your last line, dividing by $du$ is generally a big no-no. You'd also have to write $du=-\sin\theta d\theta$. $\endgroup$ – πr8 Jan 10 '16 at 10:46
  • $\begingroup$ Another way to go, is to use $u=\sqrt{9-4x^2}$. You will end up with $\int 1-\frac{9}{9-u^2}$ which will give you $u$ and logarithms after a partial fraction decomposition. $\endgroup$ – mickep Jan 10 '16 at 10:50

From where you left off, $\dfrac{\cos^2\theta}{\sin \theta}=\dfrac{1-\sin^2\theta}{\sin \theta}=\csc \theta - \sin \theta$, and $\displaystyle \int \csc \theta d\theta$ is a well known anti-derivative and you can look it up in a calculus textbook at a library near where you live!

  • $\begingroup$ so I can write the answer as $3\ln \tan \frac{\arcsin(\frac{2x}{3}}{2}+3cos(arcsin(\frac{2x}{3}))$? $\endgroup$ – gbox Jan 10 '16 at 10:49

Another way: $$I=\int\dfrac{\sqrt{9-4x^2}}x\ dx=\int\dfrac{\sqrt{9-4x^2}}{4x^2}\ 4x dx$$

Let $\sqrt{9-4x^2}=u\implies9-4x^2=u^2,4xdx=-udu$

$$I=\int\dfrac{-u^2}{9-u^2}du=\int du-9\int\dfrac{du}{9-u^2}$$

Can you take it from here?

See also: Integral of the form $a^2-x^2$


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