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$$\int_{3/4}^{3\sqrt{3}/4}\frac{x^3}{\sqrt{9-4x^2}}dx$$ So far I have the following:

$$ u=2x\Rightarrow$$

$$ u=a\sin \theta\Rightarrow 3\sin\theta$$ $$2x=3\sin\theta \rightarrow x=\frac32\sin\theta \rightarrow dx=\frac32\cos\theta d\theta$$ $$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9-9\sin^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$ $$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\sqrt{9\cos^2\theta}}\frac32\cos\theta d\theta \Rightarrow$$ $$\int_{3/4}^{3\sqrt{3}/4}\frac{u^3}{2^32} d\theta \Rightarrow$$ $$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}u^3 d\theta \Rightarrow$$ $$\frac{1}{16}\int_{3/4}^{3\sqrt{3}/4}27\sin^3\theta d\theta \Rightarrow$$ $$\frac{27}{16}\int_{3/4}^{3\sqrt{3}/4}\sin^3\theta d\theta \Rightarrow$$ $$$$ $$\int \sin^3\theta d\theta \Rightarrow \int (1-\cos^2\theta)\sin \theta d\theta$$ $$u = \cos \theta \rightarrow du=-\sin \theta d\theta \Rightarrow$$ $$\int (u^2-1)du \Rightarrow \frac{u^3}{3}-u \Rightarrow \frac{\cos^3\theta}{3}-\cos\theta $$ $$$$ $$\frac{27}{16}\bigg[\frac{\cos^3\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$ $$\frac{27}{16}\bigg[\frac{\cos^2\theta\cos\theta}{3}-\cos\theta \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$ $$\frac{27}{16}\bigg[\frac{1-4x^2\sqrt{1-\frac{4x^2}{9}}}{27}-\sqrt{1-\frac{4x^2}{9}} \bigg]\bigg|_{3/4}^{3\sqrt{3}/4}$$

Am I doing this correctly?

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    $\begingroup$ I think you forgot the square root in the denominator, and to substitute for u in the numerator. $\endgroup$ – user84413 Sep 28 '14 at 21:01
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    $\begingroup$ Now put in $u=3\sin\theta$ (and your $d\theta$), and then you are ready to integrate. [Remember that the limits you have are in terms of x.] As a check, you can also work this by letting $u=9-4x^2$, so $du=-8xdx$ and $x^2=\frac{1}{4}(9-u)$. $\endgroup$ – user84413 Sep 28 '14 at 21:58
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    $\begingroup$ No, you can't, because you are now integrating with respect to $\theta$ instead of $u$. If you wanted to, you could convert to new limits by solving $\frac{3}{2}\sin\theta=\frac{3}{4}$ and $\frac{3}{2}\sin\theta=3\sqrt{3/4}$ for $\theta$, but you could also go ahead and integrate and then substitute back to get everything in terms of x before evaluating. $\endgroup$ – user84413 Sep 28 '14 at 22:06
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    $\begingroup$ I think you want $\frac{27}{16}$ in front instead of $\frac{3}{16}$. You get $\frac{27}{8}\sin^{3}\theta$ on top, $3\cos\theta$ on the bottom, and $\frac{3}{2}\cos\theta\;d\theta$ from the $dx$. $\endgroup$ – user84413 Sep 28 '14 at 22:42
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    $\begingroup$ You are doing fine; now use that $\cos^{2}\theta=1-\sin^{2}\theta=1-(\frac{2x}{3})^2$ $\endgroup$ – user84413 Sep 28 '14 at 23:02
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You're complicating things with the double substitution. Consider $$ \sqrt{9-4x^2}=3\sqrt{1-\frac{4x^2}{9}} $$ and directly set $\frac{2}{3}x=\sin\theta$, or $x=\frac{3}{2}\sin\theta$, with $dx=\frac{3}{2}\cos\theta\,d\theta$.

If $x=\frac{3}{4}$, then you have $\frac{3}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{1}{2}$, while for $x=\frac{3\sqrt{3}}{4}$ you have $\frac{3\sqrt{3}}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{\sqrt{3}}{2}$. Thus the integral becomes $$ \int_{\pi/6}^{\pi/3}\frac{(27\sin^3\theta)/8}{3\cos\theta}\frac{3}{2}\cos\theta\,d\theta= \frac{27}{16}\int_{\pi/6}^{\pi/3}\sin^3\theta\,d\theta $$ This is computed by observing that $$ \sin^3\theta=(1-\cos^2\theta)\sin\theta $$ so you can set $\cos\theta=u$, so $-\sin\theta\,d\theta=du$ and the integral is $$ \int_{\sqrt{3}/2}^{1/2}(u^2-1)\,du $$

Your procedure is correct, but you carry around too much and doing small errors becomes easy.

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  • $\begingroup$ ahh, this is a lot easier indeed. With your above method, I've obtained: $\frac{27}{16}\bigg[\bigg(\frac{(\frac12)^3}{3}-\frac12\bigg)-\bigg(\frac{( \frac{\sqrt{3}}{2} )^3}{3}-\frac{\sqrt3}{2}\bigg)\bigg]$ Is that good? $\endgroup$ – Jessica Sep 29 '14 at 0:30
  • $\begingroup$ Reduced the above and got: $\frac{243\sqrt{3}-297}{384}$ $\endgroup$ – Jessica Sep 29 '14 at 0:49
  • $\begingroup$ Right, and you can also write this as $\frac{9}{128}(9\sqrt{3}-11)$. $\endgroup$ – user84413 Sep 29 '14 at 15:38

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