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The Integral is $$\int x^3\sqrt{9-x^2} dx$$ My method: $$x = 3\sin\theta$$ $$dx = 3\cos\theta d\theta$$ $$\sqrt{9-9\sin^2\theta} = 3\cos\theta$$ $$= \int 27\sin^3\theta (9\cos^2\theta) d\theta$$ $$=243\int (\cos^2\theta - \cos^4\theta)\sin\theta d\theta$$ $$u=\cos\theta$$ $$du=-\sin\theta d\theta$$ $$= -243\int u^2 -u^4 du$$ $$ = -81u^3 + \frac{243u^5}{5}$$ $$= -81(\cos\theta)^3 + \frac{243}{5} (\cos\theta)^5$$ $$\cos\theta = \frac{\sqrt{9-x^2}}{3}$$ $$= -27(\sqrt{9-x^2})^3 +\frac{81}{5}(\sqrt{9-x^2})^5 + C$$

Is this right? I realize that there is an easier substitution but this is how my prof wants it done. I'm confused because Wolfram Alpha gives $-\frac{1}{5}(\sqrt{9-x^2})^3(x^2+6) + C$ as the answer using the more simple substitution.

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    $\begingroup$ For future reference: use \sin and \cos, not sin and cos. $\endgroup$ – Omnomnomnom Feb 12 '14 at 2:59
  • $\begingroup$ The second u-sub is not really needed. Since the derivative of a cosine is a sine, the Chain rule has been accounted for. You could just integrate the cosines as "power-functions" $\endgroup$ – imranfat Feb 12 '14 at 3:07
  • $\begingroup$ Why do you ask whether it is right? Just differentiate the result and check... $\endgroup$ – DonAntonio Feb 12 '14 at 4:36
  • $\begingroup$ So, straightaway, $$u=\cos\theta=\sqrt{1-\frac{x^2}9}\implies 9u^2=9-x^2$$ $\endgroup$ – lab bhattacharjee Feb 12 '14 at 18:28
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Perhaps easier, shorter and/or simpler: by parts

$$u=x^2\;\;,\;\;u'=2x\\v'=x\sqrt{9-x^2}\;\;,\;\;v=-\frac13(9-x^2)^{3/2}$$

so

$$\int x^3\sqrt{9-x^2}\;dx=-\frac13x^2(9-x^2)^{3/2}+\frac23\int x(9-x^2)^{3/2}dx=$$

$$=-\frac13x^2(9-x^2)^{3/2}-\frac2{15}(9-x^2)^{5/2}+C$$

Two things: this answers negatively your last question, and we used twice above the easy-to-check fact that

$$\int f'(x)f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C\;,\;\;\forall -1\neq n\in\Bbb R$$

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