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The integral is $$\int\frac{x^3}{\sqrt{9-x^2}}dx$$ I see that if we remove the numerator I can get $\frac{1}{\sqrt{9-x^2}}$ which is the derivative of arcsin but I can't get rid of the numerator, so instead I used $$u=\sqrt{9-x^2}$$ $$du=-\frac{x}{\sqrt{9-x^2}}$$ $$u^2=9-x^2 \to u^2-9=-x^2$$ Continuing with this substitution method $$\int\frac{x^3}{u}\left(-\frac{u}{x}\right)du$$ $$\int-x^2dx \to \int(u^2-9)du=\frac{1}{3}u^3-9u+C$$ Even though this is the correct answer, I am wondering of a way to solve it that implements $\arcsin(x)$.

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  • $\begingroup$ @ShubhamJohri yep you're right, I fixed it. $\endgroup$ – Eric Brown Dec 10 '19 at 9:20
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    $\begingroup$ Have you tried $x=3\sin\theta$? $\endgroup$ – Shubham Johri Dec 10 '19 at 9:22
  • $\begingroup$ @ShubhamJohri So the way I solved it is the only U-Sub method? $\endgroup$ – Eric Brown Dec 10 '19 at 9:26
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    $\begingroup$ Read the updated comment. Another substitution could be $x^2=t$, which solves it without trigonometry. $\endgroup$ – Shubham Johri Dec 10 '19 at 9:28
  • $\begingroup$ Why not using straightforward integration by parts? Start with $$-\frac12\int x^2\frac{-2x}{\sqrt{9-x^2}}\,dx=-\frac12\int x^2\bigl(2\sqrt{9-x^2}\bigr)'\,dx.$$ $\endgroup$ – Michael Hoppe Dec 10 '19 at 10:16
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Hint:

Set $\; x=3\sin t,\enspace -\frac \pi 2<x<\frac\pi 2$, $\;\mathrm dx=3\cos t\,\mathrm d t$, to obtain $$\int\frac{x^3}{\sqrt{9-x^2}}\,\mathrm dx=\int\frac{3^3\sin^3t}{\sqrt{9(1-\sin^2t)}}\,3\cos t\,\mathrm dt=27\int\sin^3 t\,\mathrm d t=-27\int(1-\cos^2t)\, \mathrm d(\cos t).$$

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