How to calculate $int frac{x^3}{sqrt{9-x^2}}$












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According to my textbook this is : $frac{sqrt{9-x^2}^3}{3} -9 sqrt{9-x^2}$. However my solution is completely different. Here's my work:



$ x = 3sin theta, dx = 3 costheta dtheta$



$$int frac{27cos^3 theta}{sqrt{9-9sin^2theta}} = int frac{27cos^3theta}{3cos^2theta} = 9intcos^2 theta$$



$$9left(int frac{1}{2} dtheta+ int cos 2thetaright) = 9left(frac{1}{2}theta + frac{sin 2theta}{2}right)$$



Solving for $theta$ yields:



$$frac{9}{2}arcsinleft(frac{x}{3}right) + frac{9}{4}sinleft(2arcsinleft(frac{x}{3}right)right)$$



What's wrong with my solution?






calculus integration trigonometry indefinite-integrals radicals







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    where is your $mathrm dtheta$
    $endgroup$
    – ThePortakal
    Feb 2 '18 at 19:10










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    Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
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    – Crescendo
    Feb 2 '18 at 19:10










  • $begingroup$
    @Crescendo Indeed, I forgot to put the $dtheta$
    $endgroup$
    – Trey
    Feb 2 '18 at 19:13
















0












$begingroup$


According to my textbook this is : $frac{sqrt{9-x^2}^3}{3} -9 sqrt{9-x^2}$. However my solution is completely different. Here's my work:



$ x = 3sin theta, dx = 3 costheta dtheta$



$$int frac{27cos^3 theta}{sqrt{9-9sin^2theta}} = int frac{27cos^3theta}{3cos^2theta} = 9intcos^2 theta$$



$$9left(int frac{1}{2} dtheta+ int cos 2thetaright) = 9left(frac{1}{2}theta + frac{sin 2theta}{2}right)$$



Solving for $theta$ yields:



$$frac{9}{2}arcsinleft(frac{x}{3}right) + frac{9}{4}sinleft(2arcsinleft(frac{x}{3}right)right)$$



What's wrong with my solution?






calculus integration trigonometry indefinite-integrals radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    where is your $mathrm dtheta$
    $endgroup$
    – ThePortakal
    Feb 2 '18 at 19:10










  • $begingroup$
    Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
    $endgroup$
    – Crescendo
    Feb 2 '18 at 19:10










  • $begingroup$
    @Crescendo Indeed, I forgot to put the $dtheta$
    $endgroup$
    – Trey
    Feb 2 '18 at 19:13














0












0








0





$begingroup$


According to my textbook this is : $frac{sqrt{9-x^2}^3}{3} -9 sqrt{9-x^2}$. However my solution is completely different. Here's my work:



$ x = 3sin theta, dx = 3 costheta dtheta$



$$int frac{27cos^3 theta}{sqrt{9-9sin^2theta}} = int frac{27cos^3theta}{3cos^2theta} = 9intcos^2 theta$$



$$9left(int frac{1}{2} dtheta+ int cos 2thetaright) = 9left(frac{1}{2}theta + frac{sin 2theta}{2}right)$$



Solving for $theta$ yields:



$$frac{9}{2}arcsinleft(frac{x}{3}right) + frac{9}{4}sinleft(2arcsinleft(frac{x}{3}right)right)$$



What's wrong with my solution?






calculus integration trigonometry indefinite-integrals radicals







share|cite|improve this question











$endgroup$




According to my textbook this is : $frac{sqrt{9-x^2}^3}{3} -9 sqrt{9-x^2}$. However my solution is completely different. Here's my work:



$ x = 3sin theta, dx = 3 costheta dtheta$



$$int frac{27cos^3 theta}{sqrt{9-9sin^2theta}} = int frac{27cos^3theta}{3cos^2theta} = 9intcos^2 theta$$



$$9left(int frac{1}{2} dtheta+ int cos 2thetaright) = 9left(frac{1}{2}theta + frac{sin 2theta}{2}right)$$



Solving for $theta$ yields:



$$frac{9}{2}arcsinleft(frac{x}{3}right) + frac{9}{4}sinleft(2arcsinleft(frac{x}{3}right)right)$$



What's wrong with my solution?






calculus integration trigonometry indefinite-integrals radicals




calculus integration trigonometry indefinite-integrals radicals






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share|cite|improve this question













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edited Feb 2 '18 at 19:47









Michael Rozenberg

103k1891195




103k1891195










asked Feb 2 '18 at 19:03









TreyTrey

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309113












  • $begingroup$
    where is your $mathrm dtheta$
    $endgroup$
    – ThePortakal
    Feb 2 '18 at 19:10










  • $begingroup$
    Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
    $endgroup$
    – Crescendo
    Feb 2 '18 at 19:10










  • $begingroup$
    @Crescendo Indeed, I forgot to put the $dtheta$
    $endgroup$
    – Trey
    Feb 2 '18 at 19:13


















  • $begingroup$
    where is your $mathrm dtheta$
    $endgroup$
    – ThePortakal
    Feb 2 '18 at 19:10










  • $begingroup$
    Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
    $endgroup$
    – Crescendo
    Feb 2 '18 at 19:10










  • $begingroup$
    @Crescendo Indeed, I forgot to put the $dtheta$
    $endgroup$
    – Trey
    Feb 2 '18 at 19:13
















$begingroup$
where is your $mathrm dtheta$
$endgroup$
– ThePortakal
Feb 2 '18 at 19:10




$begingroup$
where is your $mathrm dtheta$
$endgroup$
– ThePortakal
Feb 2 '18 at 19:10












$begingroup$
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
$endgroup$
– Crescendo
Feb 2 '18 at 19:10




$begingroup$
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
$endgroup$
– Crescendo
Feb 2 '18 at 19:10












$begingroup$
@Crescendo Indeed, I forgot to put the $dtheta$
$endgroup$
– Trey
Feb 2 '18 at 19:13




$begingroup$
@Crescendo Indeed, I forgot to put the $dtheta$
$endgroup$
– Trey
Feb 2 '18 at 19:13










9 Answers
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You have two mistakes here.



$x = 3 sin theta$ I like that



$intfrac {x^3}{sqrt {9-x^2}} dx = int frac {27sin^3 theta}{sqrt {9-9sin^2theta}} (3cos theta) dtheta$



Your first mistake is at this step. Do you see where you went wrong?



next: $sqrt {9-9sin^2theta} = 3cos theta$ not $3cos^2 theta$



$int 27sin^3 theta dtheta$



Can you get home from here? If you are stuck try: $sin^2 theta = 1-cos^2theta$



Update:



"I still can't see how $27(-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."



$sin (arcsin frac {x}{3}) = frac {x}{3}\
sin^2 (arcsin frac {x}{3}) = frac {x^2}{9}\
1-sin^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
cos^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
cos (arcsin frac {x}{3}) = sqrt {1-frac {x^2}{9}}\
cos (arcsin frac {x}{3}) = frac {sqrt {9-x^2}}{3}\$



This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).



$27(-cos(arcsin(frac{x}{3}) + frac{cos^3(arcsin(frac{x}{3}))}{3})\
27(-sqrt {1 - frac{x^2}{9}} + frac{left(sqrt {1-frac {x^2}{9}}right)^3}{3})\
-9sqrt {9 - x^2} + frac{left(sqrt {9- x^2}right)^3}{3}
$






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  • $begingroup$
    I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
    $endgroup$
    – Trey
    Feb 2 '18 at 19:39










  • $begingroup$
    @Trey does this help?
    $endgroup$
    – Doug M
    Feb 2 '18 at 19:51










  • $begingroup$
    Yes, thank you!
    $endgroup$
    – Trey
    Feb 2 '18 at 19:55



















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$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$






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    $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
    $endgroup$
    – user477343
    Feb 4 '18 at 3:52












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    Yes, it's possible.
    $endgroup$
    – Michael Rozenberg
    Feb 4 '18 at 5:13



















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Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$






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    Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
    $endgroup$
    – user8277998
    Feb 2 '18 at 20:01



















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With your substitution we have:
$$
int frac{x^3}{sqrt{9-x^2}}dx=int frac{81sin^3 theta cos theta}{sqrt{9-9sin^2theta}}d theta
$$



that cannot be simplified as in OP.






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  • $begingroup$
    Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
    $endgroup$
    – Trey
    Feb 2 '18 at 19:18












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    Yes we can, and the integral can be simplified, but with a different result than in your question .
    $endgroup$
    – Emilio Novati
    Feb 2 '18 at 19:24





















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After doing your substitution, the numerator should become $81sin^3thetacostheta.$



Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.






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    substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be
    $$frac{1}{2}intfrac{u}{sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$






    share|cite|improve this answer









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      $$I=int frac{x^3}{sqrt{9-x^2}}dx$$
      Substitute $u=9-x^2$ then $du=-2xdx$
      $$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$






      share|cite|improve this answer









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        Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.



        $$int frac{x^3}{sqrt{9-x^2}}:dx = int x^2 cdot frac{x}{sqrt{9-x^2}}:dx $$



        Employ integration by parts:



        $$int u(x)v'(x) = u(x)v(x) - int v(x)u'(x)$$



        begin{align}
        u(x) &= x^2 & v'(x) &= frac{x}{sqrt{9 - x^2}} \
        u'(x) &= 2x & v(x) &= -sqrt{9 - x^2}
        end{align}



        Thus



        begin{align}
        int frac{x^3}{sqrt{9-x^2}}:dx &= x^2 cdot -sqrt{9 - x^2} - int -sqrt{9 - x^2} cdot 2x :dx \
        &= -x^2sqrt{9 - x^2} + 2 int xsqrt{9 - x^2}:dx \
        & = -x^2sqrt{9 - x^2} + 2 cdot -frac{1}{3}left(9 - x^2right)^{frac{3}{2}} + C \
        &= -x^2sqrt{9 - x^2}- frac{2}{3}left(9 - x^2right)^{frac{3}{2}} + C
        end{align}



        Where $C$ is a constant of integration






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          1












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          Why not one more method?



          begin{align}
          int frac{x^3}{sqrt{9 - x^2}}:dx &= int x cdot frac{x^2}{sqrt{9 - x^2}}:dx = - int x cdot frac{-x^2}{sqrt{9 - x^2}}:dx \
          &=- int x cdot frac{-x^2 - 9 + 9}{sqrt{9 - x^2}}:dx = -int x cdot left[ frac{9 - x^2}{sqrt{9 -x^2}} - frac{9}{sqrt{9 - x^2}} right]:dx\
          &= -int xsqrt{9 = x^2}:dx + int frac{9x}{sqrt{9 - x^2}}:dx \
          &= frac{1}{3}left(9 - x^2right)^{frac{3}{2}} - 9sqrt{9 - x^2} + C
          end{align}



          Where $C$ is a constant of integration.






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            9 Answers
            9






            active

            oldest

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            9 Answers
            9






            active

            oldest

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            You have two mistakes here.



            $x = 3 sin theta$ I like that



            $intfrac {x^3}{sqrt {9-x^2}} dx = int frac {27sin^3 theta}{sqrt {9-9sin^2theta}} (3cos theta) dtheta$



            Your first mistake is at this step. Do you see where you went wrong?



            next: $sqrt {9-9sin^2theta} = 3cos theta$ not $3cos^2 theta$



            $int 27sin^3 theta dtheta$



            Can you get home from here? If you are stuck try: $sin^2 theta = 1-cos^2theta$



            Update:



            "I still can't see how $27(-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."



            $sin (arcsin frac {x}{3}) = frac {x}{3}\
            sin^2 (arcsin frac {x}{3}) = frac {x^2}{9}\
            1-sin^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos (arcsin frac {x}{3}) = sqrt {1-frac {x^2}{9}}\
            cos (arcsin frac {x}{3}) = frac {sqrt {9-x^2}}{3}\$



            This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).



            $27(-cos(arcsin(frac{x}{3}) + frac{cos^3(arcsin(frac{x}{3}))}{3})\
            27(-sqrt {1 - frac{x^2}{9}} + frac{left(sqrt {1-frac {x^2}{9}}right)^3}{3})\
            -9sqrt {9 - x^2} + frac{left(sqrt {9- x^2}right)^3}{3}
            $






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
              $endgroup$
              – Trey
              Feb 2 '18 at 19:39










            • $begingroup$
              @Trey does this help?
              $endgroup$
              – Doug M
              Feb 2 '18 at 19:51










            • $begingroup$
              Yes, thank you!
              $endgroup$
              – Trey
              Feb 2 '18 at 19:55
















            1












            $begingroup$

            You have two mistakes here.



            $x = 3 sin theta$ I like that



            $intfrac {x^3}{sqrt {9-x^2}} dx = int frac {27sin^3 theta}{sqrt {9-9sin^2theta}} (3cos theta) dtheta$



            Your first mistake is at this step. Do you see where you went wrong?



            next: $sqrt {9-9sin^2theta} = 3cos theta$ not $3cos^2 theta$



            $int 27sin^3 theta dtheta$



            Can you get home from here? If you are stuck try: $sin^2 theta = 1-cos^2theta$



            Update:



            "I still can't see how $27(-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."



            $sin (arcsin frac {x}{3}) = frac {x}{3}\
            sin^2 (arcsin frac {x}{3}) = frac {x^2}{9}\
            1-sin^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos (arcsin frac {x}{3}) = sqrt {1-frac {x^2}{9}}\
            cos (arcsin frac {x}{3}) = frac {sqrt {9-x^2}}{3}\$



            This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).



            $27(-cos(arcsin(frac{x}{3}) + frac{cos^3(arcsin(frac{x}{3}))}{3})\
            27(-sqrt {1 - frac{x^2}{9}} + frac{left(sqrt {1-frac {x^2}{9}}right)^3}{3})\
            -9sqrt {9 - x^2} + frac{left(sqrt {9- x^2}right)^3}{3}
            $






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
              $endgroup$
              – Trey
              Feb 2 '18 at 19:39










            • $begingroup$
              @Trey does this help?
              $endgroup$
              – Doug M
              Feb 2 '18 at 19:51










            • $begingroup$
              Yes, thank you!
              $endgroup$
              – Trey
              Feb 2 '18 at 19:55














            1












            1








            1





            $begingroup$

            You have two mistakes here.



            $x = 3 sin theta$ I like that



            $intfrac {x^3}{sqrt {9-x^2}} dx = int frac {27sin^3 theta}{sqrt {9-9sin^2theta}} (3cos theta) dtheta$



            Your first mistake is at this step. Do you see where you went wrong?



            next: $sqrt {9-9sin^2theta} = 3cos theta$ not $3cos^2 theta$



            $int 27sin^3 theta dtheta$



            Can you get home from here? If you are stuck try: $sin^2 theta = 1-cos^2theta$



            Update:



            "I still can't see how $27(-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."



            $sin (arcsin frac {x}{3}) = frac {x}{3}\
            sin^2 (arcsin frac {x}{3}) = frac {x^2}{9}\
            1-sin^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos (arcsin frac {x}{3}) = sqrt {1-frac {x^2}{9}}\
            cos (arcsin frac {x}{3}) = frac {sqrt {9-x^2}}{3}\$



            This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).



            $27(-cos(arcsin(frac{x}{3}) + frac{cos^3(arcsin(frac{x}{3}))}{3})\
            27(-sqrt {1 - frac{x^2}{9}} + frac{left(sqrt {1-frac {x^2}{9}}right)^3}{3})\
            -9sqrt {9 - x^2} + frac{left(sqrt {9- x^2}right)^3}{3}
            $






            share|cite|improve this answer











            $endgroup$



            You have two mistakes here.



            $x = 3 sin theta$ I like that



            $intfrac {x^3}{sqrt {9-x^2}} dx = int frac {27sin^3 theta}{sqrt {9-9sin^2theta}} (3cos theta) dtheta$



            Your first mistake is at this step. Do you see where you went wrong?



            next: $sqrt {9-9sin^2theta} = 3cos theta$ not $3cos^2 theta$



            $int 27sin^3 theta dtheta$



            Can you get home from here? If you are stuck try: $sin^2 theta = 1-cos^2theta$



            Update:



            "I still can't see how $27(-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3})$ is equivalent to the solution given in the textbook."



            $sin (arcsin frac {x}{3}) = frac {x}{3}\
            sin^2 (arcsin frac {x}{3}) = frac {x^2}{9}\
            1-sin^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos^2 (arcsin frac {x}{3}) = 1-frac {x^2}{9}\
            cos (arcsin frac {x}{3}) = sqrt {1-frac {x^2}{9}}\
            cos (arcsin frac {x}{3}) = frac {sqrt {9-x^2}}{3}\$



            This can also be done geometrically, and I suggest you prove it this way (an exercise left to the reader).



            $27(-cos(arcsin(frac{x}{3}) + frac{cos^3(arcsin(frac{x}{3}))}{3})\
            27(-sqrt {1 - frac{x^2}{9}} + frac{left(sqrt {1-frac {x^2}{9}}right)^3}{3})\
            -9sqrt {9 - x^2} + frac{left(sqrt {9- x^2}right)^3}{3}
            $







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            edited Feb 2 '18 at 19:51

























            answered Feb 2 '18 at 19:16









            Doug MDoug M

            45.1k31854




            45.1k31854












            • $begingroup$
              I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
              $endgroup$
              – Trey
              Feb 2 '18 at 19:39










            • $begingroup$
              @Trey does this help?
              $endgroup$
              – Doug M
              Feb 2 '18 at 19:51










            • $begingroup$
              Yes, thank you!
              $endgroup$
              – Trey
              Feb 2 '18 at 19:55


















            • $begingroup$
              I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
              $endgroup$
              – Trey
              Feb 2 '18 at 19:39










            • $begingroup$
              @Trey does this help?
              $endgroup$
              – Doug M
              Feb 2 '18 at 19:51










            • $begingroup$
              Yes, thank you!
              $endgroup$
              – Trey
              Feb 2 '18 at 19:55
















            $begingroup$
            I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
            $endgroup$
            – Trey
            Feb 2 '18 at 19:39




            $begingroup$
            I still can't see how $-cos(arcsin(frac{x}{3})) + frac{cos^3(arcsin(frac{x}{3}))}{3}$ is equivalent to the solution given in the textbook
            $endgroup$
            – Trey
            Feb 2 '18 at 19:39












            $begingroup$
            @Trey does this help?
            $endgroup$
            – Doug M
            Feb 2 '18 at 19:51




            $begingroup$
            @Trey does this help?
            $endgroup$
            – Doug M
            Feb 2 '18 at 19:51












            $begingroup$
            Yes, thank you!
            $endgroup$
            – Trey
            Feb 2 '18 at 19:55




            $begingroup$
            Yes, thank you!
            $endgroup$
            – Trey
            Feb 2 '18 at 19:55











            3












            $begingroup$

            $$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
            $$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
              $endgroup$
              – user477343
              Feb 4 '18 at 3:52












            • $begingroup$
              Yes, it's possible.
              $endgroup$
              – Michael Rozenberg
              Feb 4 '18 at 5:13
















            3












            $begingroup$

            $$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
            $$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
              $endgroup$
              – user477343
              Feb 4 '18 at 3:52












            • $begingroup$
              Yes, it's possible.
              $endgroup$
              – Michael Rozenberg
              Feb 4 '18 at 5:13














            3












            3








            3





            $begingroup$

            $$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
            $$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$






            share|cite|improve this answer









            $endgroup$



            $$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
            $$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 2 '18 at 19:45









            Michael RozenbergMichael Rozenberg

            103k1891195




            103k1891195












            • $begingroup$
              $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
              $endgroup$
              – user477343
              Feb 4 '18 at 3:52












            • $begingroup$
              Yes, it's possible.
              $endgroup$
              – Michael Rozenberg
              Feb 4 '18 at 5:13


















            • $begingroup$
              $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
              $endgroup$
              – user477343
              Feb 4 '18 at 3:52












            • $begingroup$
              Yes, it's possible.
              $endgroup$
              – Michael Rozenberg
              Feb 4 '18 at 5:13
















            $begingroup$
            $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
            $endgroup$
            – user477343
            Feb 4 '18 at 3:52






            $begingroup$
            $$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
            $endgroup$
            – user477343
            Feb 4 '18 at 3:52














            $begingroup$
            Yes, it's possible.
            $endgroup$
            – Michael Rozenberg
            Feb 4 '18 at 5:13




            $begingroup$
            Yes, it's possible.
            $endgroup$
            – Michael Rozenberg
            Feb 4 '18 at 5:13











            2












            $begingroup$

            Let $u = 9-x^2$, $dx=frac{du}{-2x}$
            $$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
              $endgroup$
              – user8277998
              Feb 2 '18 at 20:01
















            2












            $begingroup$

            Let $u = 9-x^2$, $dx=frac{du}{-2x}$
            $$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
              $endgroup$
              – user8277998
              Feb 2 '18 at 20:01














            2












            2








            2





            $begingroup$

            Let $u = 9-x^2$, $dx=frac{du}{-2x}$
            $$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$






            share|cite|improve this answer









            $endgroup$



            Let $u = 9-x^2$, $dx=frac{du}{-2x}$
            $$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 2 '18 at 19:13









            id500id500

            325111




            325111












            • $begingroup$
              Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
              $endgroup$
              – user8277998
              Feb 2 '18 at 20:01


















            • $begingroup$
              Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
              $endgroup$
              – user8277998
              Feb 2 '18 at 20:01
















            $begingroup$
            Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
            $endgroup$
            – user8277998
            Feb 2 '18 at 20:01




            $begingroup$
            Great answer. I don't understand why people are so stoked about trig substitutions and use for really basic integrals. Trig substitutions are such a mess, agh !
            $endgroup$
            – user8277998
            Feb 2 '18 at 20:01











            2












            $begingroup$

            With your substitution we have:
            $$
            int frac{x^3}{sqrt{9-x^2}}dx=int frac{81sin^3 theta cos theta}{sqrt{9-9sin^2theta}}d theta
            $$



            that cannot be simplified as in OP.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
              $endgroup$
              – Trey
              Feb 2 '18 at 19:18












            • $begingroup$
              Yes we can, and the integral can be simplified, but with a different result than in your question .
              $endgroup$
              – Emilio Novati
              Feb 2 '18 at 19:24


















            2












            $begingroup$

            With your substitution we have:
            $$
            int frac{x^3}{sqrt{9-x^2}}dx=int frac{81sin^3 theta cos theta}{sqrt{9-9sin^2theta}}d theta
            $$



            that cannot be simplified as in OP.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
              $endgroup$
              – Trey
              Feb 2 '18 at 19:18












            • $begingroup$
              Yes we can, and the integral can be simplified, but with a different result than in your question .
              $endgroup$
              – Emilio Novati
              Feb 2 '18 at 19:24
















            2












            2








            2





            $begingroup$

            With your substitution we have:
            $$
            int frac{x^3}{sqrt{9-x^2}}dx=int frac{81sin^3 theta cos theta}{sqrt{9-9sin^2theta}}d theta
            $$



            that cannot be simplified as in OP.






            share|cite|improve this answer











            $endgroup$



            With your substitution we have:
            $$
            int frac{x^3}{sqrt{9-x^2}}dx=int frac{81sin^3 theta cos theta}{sqrt{9-9sin^2theta}}d theta
            $$



            that cannot be simplified as in OP.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 2 '18 at 19:23

























            answered Feb 2 '18 at 19:10









            Emilio NovatiEmilio Novati

            52k43474




            52k43474












            • $begingroup$
              Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
              $endgroup$
              – Trey
              Feb 2 '18 at 19:18












            • $begingroup$
              Yes we can, and the integral can be simplified, but with a different result than in your question .
              $endgroup$
              – Emilio Novati
              Feb 2 '18 at 19:24




















            • $begingroup$
              Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
              $endgroup$
              – Trey
              Feb 2 '18 at 19:18












            • $begingroup$
              Yes we can, and the integral can be simplified, but with a different result than in your question .
              $endgroup$
              – Emilio Novati
              Feb 2 '18 at 19:24


















            $begingroup$
            Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
            $endgroup$
            – Trey
            Feb 2 '18 at 19:18






            $begingroup$
            Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
            $endgroup$
            – Trey
            Feb 2 '18 at 19:18














            $begingroup$
            Yes we can, and the integral can be simplified, but with a different result than in your question .
            $endgroup$
            – Emilio Novati
            Feb 2 '18 at 19:24






            $begingroup$
            Yes we can, and the integral can be simplified, but with a different result than in your question .
            $endgroup$
            – Emilio Novati
            Feb 2 '18 at 19:24













            1












            $begingroup$

            After doing your substitution, the numerator should become $81sin^3thetacostheta.$



            Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              After doing your substitution, the numerator should become $81sin^3thetacostheta.$



              Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                After doing your substitution, the numerator should become $81sin^3thetacostheta.$



                Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.






                share|cite|improve this answer









                $endgroup$



                After doing your substitution, the numerator should become $81sin^3thetacostheta.$



                Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 2 '18 at 19:09









                José Carlos SantosJosé Carlos Santos

                161k22127232




                161k22127232























                    1












                    $begingroup$

                    substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be
                    $$frac{1}{2}intfrac{u}{sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$






                    share|cite|improve this answer









                    $endgroup$


















                      1












                      $begingroup$

                      substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be
                      $$frac{1}{2}intfrac{u}{sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$






                      share|cite|improve this answer









                      $endgroup$
















                        1












                        1








                        1





                        $begingroup$

                        substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be
                        $$frac{1}{2}intfrac{u}{sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$






                        share|cite|improve this answer









                        $endgroup$



                        substituting $$u=x^2$$ then we have $$du=2xdx$$ and our integral will be
                        $$frac{1}{2}intfrac{u}{sqrt{9-u}}du$$ no let $$s=9-u$$ and with $$ds=-du$$our integral is $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Feb 2 '18 at 19:10









                        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                        75.3k42865




                        75.3k42865























                            1












                            $begingroup$

                            $$I=int frac{x^3}{sqrt{9-x^2}}dx$$
                            Substitute $u=9-x^2$ then $du=-2xdx$
                            $$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$






                            share|cite|improve this answer









                            $endgroup$


















                              1












                              $begingroup$

                              $$I=int frac{x^3}{sqrt{9-x^2}}dx$$
                              Substitute $u=9-x^2$ then $du=-2xdx$
                              $$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$






                              share|cite|improve this answer









                              $endgroup$
















                                1












                                1








                                1





                                $begingroup$

                                $$I=int frac{x^3}{sqrt{9-x^2}}dx$$
                                Substitute $u=9-x^2$ then $du=-2xdx$
                                $$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$






                                share|cite|improve this answer









                                $endgroup$



                                $$I=int frac{x^3}{sqrt{9-x^2}}dx$$
                                Substitute $u=9-x^2$ then $du=-2xdx$
                                $$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Feb 2 '18 at 19:28









                                IshamIsham

                                12.7k3929




                                12.7k3929























                                    1












                                    $begingroup$

                                    Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.



                                    $$int frac{x^3}{sqrt{9-x^2}}:dx = int x^2 cdot frac{x}{sqrt{9-x^2}}:dx $$



                                    Employ integration by parts:



                                    $$int u(x)v'(x) = u(x)v(x) - int v(x)u'(x)$$



                                    begin{align}
                                    u(x) &= x^2 & v'(x) &= frac{x}{sqrt{9 - x^2}} \
                                    u'(x) &= 2x & v(x) &= -sqrt{9 - x^2}
                                    end{align}



                                    Thus



                                    begin{align}
                                    int frac{x^3}{sqrt{9-x^2}}:dx &= x^2 cdot -sqrt{9 - x^2} - int -sqrt{9 - x^2} cdot 2x :dx \
                                    &= -x^2sqrt{9 - x^2} + 2 int xsqrt{9 - x^2}:dx \
                                    & = -x^2sqrt{9 - x^2} + 2 cdot -frac{1}{3}left(9 - x^2right)^{frac{3}{2}} + C \
                                    &= -x^2sqrt{9 - x^2}- frac{2}{3}left(9 - x^2right)^{frac{3}{2}} + C
                                    end{align}



                                    Where $C$ is a constant of integration






                                    share|cite|improve this answer









                                    $endgroup$


















                                      1












                                      $begingroup$

                                      Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.



                                      $$int frac{x^3}{sqrt{9-x^2}}:dx = int x^2 cdot frac{x}{sqrt{9-x^2}}:dx $$



                                      Employ integration by parts:



                                      $$int u(x)v'(x) = u(x)v(x) - int v(x)u'(x)$$



                                      begin{align}
                                      u(x) &= x^2 & v'(x) &= frac{x}{sqrt{9 - x^2}} \
                                      u'(x) &= 2x & v(x) &= -sqrt{9 - x^2}
                                      end{align}



                                      Thus



                                      begin{align}
                                      int frac{x^3}{sqrt{9-x^2}}:dx &= x^2 cdot -sqrt{9 - x^2} - int -sqrt{9 - x^2} cdot 2x :dx \
                                      &= -x^2sqrt{9 - x^2} + 2 int xsqrt{9 - x^2}:dx \
                                      & = -x^2sqrt{9 - x^2} + 2 cdot -frac{1}{3}left(9 - x^2right)^{frac{3}{2}} + C \
                                      &= -x^2sqrt{9 - x^2}- frac{2}{3}left(9 - x^2right)^{frac{3}{2}} + C
                                      end{align}



                                      Where $C$ is a constant of integration






                                      share|cite|improve this answer









                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.



                                        $$int frac{x^3}{sqrt{9-x^2}}:dx = int x^2 cdot frac{x}{sqrt{9-x^2}}:dx $$



                                        Employ integration by parts:



                                        $$int u(x)v'(x) = u(x)v(x) - int v(x)u'(x)$$



                                        begin{align}
                                        u(x) &= x^2 & v'(x) &= frac{x}{sqrt{9 - x^2}} \
                                        u'(x) &= 2x & v(x) &= -sqrt{9 - x^2}
                                        end{align}



                                        Thus



                                        begin{align}
                                        int frac{x^3}{sqrt{9-x^2}}:dx &= x^2 cdot -sqrt{9 - x^2} - int -sqrt{9 - x^2} cdot 2x :dx \
                                        &= -x^2sqrt{9 - x^2} + 2 int xsqrt{9 - x^2}:dx \
                                        & = -x^2sqrt{9 - x^2} + 2 cdot -frac{1}{3}left(9 - x^2right)^{frac{3}{2}} + C \
                                        &= -x^2sqrt{9 - x^2}- frac{2}{3}left(9 - x^2right)^{frac{3}{2}} + C
                                        end{align}



                                        Where $C$ is a constant of integration






                                        share|cite|improve this answer









                                        $endgroup$



                                        Plenty of good answers, but as the expression goes 'There's more than one way to skin a Cat' so I will provide another.



                                        $$int frac{x^3}{sqrt{9-x^2}}:dx = int x^2 cdot frac{x}{sqrt{9-x^2}}:dx $$



                                        Employ integration by parts:



                                        $$int u(x)v'(x) = u(x)v(x) - int v(x)u'(x)$$



                                        begin{align}
                                        u(x) &= x^2 & v'(x) &= frac{x}{sqrt{9 - x^2}} \
                                        u'(x) &= 2x & v(x) &= -sqrt{9 - x^2}
                                        end{align}



                                        Thus



                                        begin{align}
                                        int frac{x^3}{sqrt{9-x^2}}:dx &= x^2 cdot -sqrt{9 - x^2} - int -sqrt{9 - x^2} cdot 2x :dx \
                                        &= -x^2sqrt{9 - x^2} + 2 int xsqrt{9 - x^2}:dx \
                                        & = -x^2sqrt{9 - x^2} + 2 cdot -frac{1}{3}left(9 - x^2right)^{frac{3}{2}} + C \
                                        &= -x^2sqrt{9 - x^2}- frac{2}{3}left(9 - x^2right)^{frac{3}{2}} + C
                                        end{align}



                                        Where $C$ is a constant of integration







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 8 '18 at 7:03









                                        DavidGDavidG

                                        2,0821723




                                        2,0821723























                                            1












                                            $begingroup$

                                            Why not one more method?



                                            begin{align}
                                            int frac{x^3}{sqrt{9 - x^2}}:dx &= int x cdot frac{x^2}{sqrt{9 - x^2}}:dx = - int x cdot frac{-x^2}{sqrt{9 - x^2}}:dx \
                                            &=- int x cdot frac{-x^2 - 9 + 9}{sqrt{9 - x^2}}:dx = -int x cdot left[ frac{9 - x^2}{sqrt{9 -x^2}} - frac{9}{sqrt{9 - x^2}} right]:dx\
                                            &= -int xsqrt{9 = x^2}:dx + int frac{9x}{sqrt{9 - x^2}}:dx \
                                            &= frac{1}{3}left(9 - x^2right)^{frac{3}{2}} - 9sqrt{9 - x^2} + C
                                            end{align}



                                            Where $C$ is a constant of integration.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              Why not one more method?



                                              begin{align}
                                              int frac{x^3}{sqrt{9 - x^2}}:dx &= int x cdot frac{x^2}{sqrt{9 - x^2}}:dx = - int x cdot frac{-x^2}{sqrt{9 - x^2}}:dx \
                                              &=- int x cdot frac{-x^2 - 9 + 9}{sqrt{9 - x^2}}:dx = -int x cdot left[ frac{9 - x^2}{sqrt{9 -x^2}} - frac{9}{sqrt{9 - x^2}} right]:dx\
                                              &= -int xsqrt{9 = x^2}:dx + int frac{9x}{sqrt{9 - x^2}}:dx \
                                              &= frac{1}{3}left(9 - x^2right)^{frac{3}{2}} - 9sqrt{9 - x^2} + C
                                              end{align}



                                              Where $C$ is a constant of integration.






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                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$

                                                Why not one more method?



                                                begin{align}
                                                int frac{x^3}{sqrt{9 - x^2}}:dx &= int x cdot frac{x^2}{sqrt{9 - x^2}}:dx = - int x cdot frac{-x^2}{sqrt{9 - x^2}}:dx \
                                                &=- int x cdot frac{-x^2 - 9 + 9}{sqrt{9 - x^2}}:dx = -int x cdot left[ frac{9 - x^2}{sqrt{9 -x^2}} - frac{9}{sqrt{9 - x^2}} right]:dx\
                                                &= -int xsqrt{9 = x^2}:dx + int frac{9x}{sqrt{9 - x^2}}:dx \
                                                &= frac{1}{3}left(9 - x^2right)^{frac{3}{2}} - 9sqrt{9 - x^2} + C
                                                end{align}



                                                Where $C$ is a constant of integration.






                                                share|cite|improve this answer









                                                $endgroup$



                                                Why not one more method?



                                                begin{align}
                                                int frac{x^3}{sqrt{9 - x^2}}:dx &= int x cdot frac{x^2}{sqrt{9 - x^2}}:dx = - int x cdot frac{-x^2}{sqrt{9 - x^2}}:dx \
                                                &=- int x cdot frac{-x^2 - 9 + 9}{sqrt{9 - x^2}}:dx = -int x cdot left[ frac{9 - x^2}{sqrt{9 -x^2}} - frac{9}{sqrt{9 - x^2}} right]:dx\
                                                &= -int xsqrt{9 = x^2}:dx + int frac{9x}{sqrt{9 - x^2}}:dx \
                                                &= frac{1}{3}left(9 - x^2right)^{frac{3}{2}} - 9sqrt{9 - x^2} + C
                                                end{align}



                                                Where $C$ is a constant of integration.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Dec 8 '18 at 8:51









                                                DavidGDavidG

                                                2,0821723




                                                2,0821723






























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