How to calculate $int frac{x^3}{sqrt{9-x^2}}$
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
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
add a comment |
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
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13
add a comment |
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
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
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
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
edited Feb 2 at 19:47
Michael Rozenberg
95.4k1588184
95.4k1588184
asked Feb 2 at 19:03
Trey
304113
asked Feb 2 at 19:03
Trey
304113
asked Feb 2 at 19:03
Trey
304113
304113
where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13
add a comment |
where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13
where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13
add a comment |
9 Answers
9
active
oldest
votes
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}
$
answered Feb 2 at 19:16
Doug M
43.9k31854
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
add a comment |
$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
add a comment |
Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$
answered Feb 2 at 19:13
id500
32511
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 !
– user8277998
Feb 2 at 20:01
add a comment |
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.
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
add a comment |
After doing your substitution, the numerator should become $81sin^3thetacostheta.$
Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
add a comment |
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 $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
add a comment |
$$I=int frac{x^3}{sqrt{9-x^2}}dx$$
Substitute $u=9-x^2$ then $du=-2xdx$
$$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$
answered Feb 2 at 19:28
Isham
12.7k3929
add a comment |
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
answered Dec 8 at 7:03
DavidG
1,561619
add a comment |
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.
answered Dec 8 at 8:51
DavidG
1,561619
add a comment |
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9 Answers
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9 Answers
9
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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}
$
answered Feb 2 at 19:16
Doug M
43.9k31854
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
add a comment |
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}
$
answered Feb 2 at 19:16
Doug M
43.9k31854
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
add a comment |
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}
$
answered Feb 2 at 19:16
Doug M
43.9k31854
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}
$
answered Feb 2 at 19:16
Doug M
43.9k31854
edited Feb 2 at 19:51
answered Feb 2 at 19:16
Doug M
43.9k31854
answered Feb 2 at 19:16
Doug M
43.9k31854
answered Feb 2 at 19:16
Doug M
43.9k31854
43.9k31854
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
add a comment |
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
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
– Trey
Feb 2 at 19:39
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
– Trey
Feb 2 at 19:39
@Trey does this help?
– Doug M
Feb 2 at 19:51
@Trey does this help?
– Doug M
Feb 2 at 19:51
Yes, thank you!
– Trey
Feb 2 at 19:55
Yes, thank you!
– Trey
Feb 2 at 19:55
add a comment |
$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
add a comment |
$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
add a comment |
$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
$$intfrac{x^3-9x+9x}{sqrt{9-x^2}}dx=-int xsqrt{9-x^2}dx+9intfrac{x}{sqrt{9-x^2}}dx=$$
$$=frac{1}{3}sqrt{(9-x^2)^3}-9sqrt{9-x^2}+C.$$
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
answered Feb 2 at 19:45
Michael Rozenberg
95.4k1588184
95.4k1588184
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
add a comment |
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
$$= sqrt{9 - x^2}bigg(frac 13left(9 - x^2right) - 9bigg) + C = -sqrt{9 - x^2}bigg(6 + frac{x^2}{3}bigg) + C$$ I think.
– user477343
Feb 4 at 3:52
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
Yes, it's possible.
– Michael Rozenberg
Feb 4 at 5:13
add a comment |
Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$
answered Feb 2 at 19:13
id500
32511
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 !
– user8277998
Feb 2 at 20:01
add a comment |
Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$
answered Feb 2 at 19:13
id500
32511
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 !
– user8277998
Feb 2 at 20:01
add a comment |
Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$
answered Feb 2 at 19:13
id500
32511
Let $u = 9-x^2$, $dx=frac{du}{-2x}$
$$int frac{x^3}{sqrt{9-x^2}}dx=-1/2int frac{9-u}{sqrt{u}}du=-1/2[9(2)u^{1/2}-(2/3)u^{3/2}]\=-9(9-x^2)^{1/2}+1/3(9-x^2)^{1/3}$$
answered Feb 2 at 19:13
id500
32511
answered Feb 2 at 19:13
id500
32511
answered Feb 2 at 19:13
id500
32511
answered Feb 2 at 19:13
id500
32511
32511
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 !
– user8277998
Feb 2 at 20:01
add a comment |
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 !
– user8277998
Feb 2 at 20:01
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 !
– user8277998
Feb 2 at 20:01
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 !
– user8277998
Feb 2 at 20:01
add a comment |
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.
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
add a comment |
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.
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
add a comment |
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.
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
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.
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
edited Feb 2 at 19:23
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
answered Feb 2 at 19:10
Emilio Novati
51.5k43472
51.5k43472
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
add a comment |
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Can't I rewrite the denominator as $sqrt{9(1-sin^2theta)}$ which in turn gives ${3costheta}$?
– Trey
Feb 2 at 19:18
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
Yes we can, and the integral can be simplified, but with a different result than in your question .
– Emilio Novati
Feb 2 at 19:24
add a comment |
After doing your substitution, the numerator should become $81sin^3thetacostheta.$
Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
add a comment |
After doing your substitution, the numerator should become $81sin^3thetacostheta.$
Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
add a comment |
After doing your substitution, the numerator should become $81sin^3thetacostheta.$
Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
After doing your substitution, the numerator should become $81sin^3thetacostheta.$
Anyway, my suggestion is that you do $x^2=y$ and $2x,mathrm dx=mathrm dy$.
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
answered Feb 2 at 19:09
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
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 $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
add a comment |
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 $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
add a comment |
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 $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
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 $$-frac{1}{2}intleft(frac{9}{sqrt{s}}-sqrt{s}right)ds$$
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
answered Feb 2 at 19:10
Dr. Sonnhard Graubner
72.8k32865
72.8k32865
add a comment |
add a comment |
$$I=int frac{x^3}{sqrt{9-x^2}}dx$$
Substitute $u=9-x^2$ then $du=-2xdx$
$$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$
answered Feb 2 at 19:28
Isham
12.7k3929
add a comment |
$$I=int frac{x^3}{sqrt{9-x^2}}dx$$
Substitute $u=9-x^2$ then $du=-2xdx$
$$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$
answered Feb 2 at 19:28
Isham
12.7k3929
add a comment |
$$I=int frac{x^3}{sqrt{9-x^2}}dx$$
Substitute $u=9-x^2$ then $du=-2xdx$
$$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$
answered Feb 2 at 19:28
Isham
12.7k3929
$$I=int frac{x^3}{sqrt{9-x^2}}dx$$
Substitute $u=9-x^2$ then $du=-2xdx$
$$I=int frac{x^3}{sqrt{9-x^2}}dx=-frac12int frac {9-u}{sqrt u}du$$
answered Feb 2 at 19:28
Isham
12.7k3929
answered Feb 2 at 19:28
Isham
12.7k3929
answered Feb 2 at 19:28
Isham
12.7k3929
answered Feb 2 at 19:28
Isham
12.7k3929
12.7k3929
add a comment |
add a comment |
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
answered Dec 8 at 7:03
DavidG
1,561619
add a comment |
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
answered Dec 8 at 7:03
DavidG
1,561619
add a comment |
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
answered Dec 8 at 7:03
DavidG
1,561619
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
answered Dec 8 at 7:03
DavidG
1,561619
answered Dec 8 at 7:03
DavidG
1,561619
answered Dec 8 at 7:03
DavidG
1,561619
answered Dec 8 at 7:03
DavidG
1,561619
1,561619
add a comment |
add a comment |
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.
answered Dec 8 at 8:51
DavidG
1,561619
add a comment |
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.
answered Dec 8 at 8:51
DavidG
1,561619
add a comment |
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.
answered Dec 8 at 8:51
DavidG
1,561619
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.
answered Dec 8 at 8:51
DavidG
1,561619
answered Dec 8 at 8:51
DavidG
1,561619
answered Dec 8 at 8:51
DavidG
1,561619
answered Dec 8 at 8:51
DavidG
1,561619
1,561619
add a comment |
add a comment |
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where is your $mathrm dtheta$
– ThePortakal
Feb 2 at 19:10
Shouldn't your integral be$$int dtheta,frac {81sin^3thetacostheta}{sqrt{9-9sin^2theta}}$$
– Crescendo
Feb 2 at 19:10
@Crescendo Indeed, I forgot to put the $dtheta$
– Trey
Feb 2 at 19:13