Mixing
A reduction formula for $int_0^1 x^n/sqrt{9 - x^2},mathrm dx$
$begingroup$
Let $$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2sqrt2$$
I've found that $displaystyle I_0 = sin^{-1}left(frac{1}{3}right)$, but that's it.
I'm struggling to go any further. Anyone have any hints?
calculus integration definite-integrals radicals reduction-formula
edited Jan 25 at 2:36
clathratus
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
$endgroup$
add a comment |
$begingroup$
Let $$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2sqrt2$$
I've found that $displaystyle I_0 = sin^{-1}left(frac{1}{3}right)$, but that's it.
I'm struggling to go any further. Anyone have any hints?
calculus integration definite-integrals radicals reduction-formula
edited Jan 25 at 2:36
clathratus
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
$endgroup$
$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51
add a comment |
$begingroup$
Let $$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2sqrt2$$
I've found that $displaystyle I_0 = sin^{-1}left(frac{1}{3}right)$, but that's it.
I'm struggling to go any further. Anyone have any hints?
calculus integration definite-integrals radicals reduction-formula
edited Jan 25 at 2:36
clathratus
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
$endgroup$
Let $$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx$$
Using integration, show that $$nI_n = 9(n - 1)nI_{n - 2} - 2sqrt2$$
I've found that $displaystyle I_0 = sin^{-1}left(frac{1}{3}right)$, but that's it.
I'm struggling to go any further. Anyone have any hints?
calculus integration definite-integrals radicals reduction-formula
calculus integration definite-integrals radicals reduction-formula
edited Jan 25 at 2:36
clathratus
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
edited Jan 25 at 2:36
clathratus
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
edited Jan 25 at 2:36
clathratus
5,0351338
edited Jan 25 at 2:36
clathratus
5,0351338
edited Jan 25 at 2:36
clathratus
5,0351338
5,0351338
asked Dec 14 '14 at 0:07
jane brownjane brown
283
asked Dec 14 '14 at 0:07
jane brownjane brown
283
asked Dec 14 '14 at 0:07
jane brownjane brown
283
283
$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51
add a comment |
$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51
$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
Using Integration By Parts
$$frac{x}{sqrt{9 - x^2}},mathrm dx=,mathrm dviff-sqrt{9-x^2}=v$$
$$x^{n-1}=uiff (n-1)x^{n-2},mathrm dx=,mathrm du$$
$$begin{align}
I_n
&= -x^{n-1}cdotsqrt{9-x^2}Bigg|_0^1+(n-1)int_0^1x^{n-2}sqrt{9-x^2},mathrm dxtag{1}\
&= -2sqrt2+(n-1)int_0^1frac{x^{n-2}(9-x^2)}{sqrt{9-x^2}},mathrm dxtag{2}\
&= -2sqrt2+9(n-1)int_0^1frac{x^{n-2}}{sqrt{9-x^2}},mathrm dx-(n-1)int_0^1frac{x^{n}}{sqrt{9-x^2}},mathrm dxtag{3}\
&= -2sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}tag{4}\
end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$
edited Jul 9 '18 at 17:51
tatan
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
$endgroup$
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
add a comment |
$begingroup$
Hint:-
$x=3sin theta implies dx=3costheta dtheta$
$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
$endgroup$
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
Using Integration By Parts
$$frac{x}{sqrt{9 - x^2}},mathrm dx=,mathrm dviff-sqrt{9-x^2}=v$$
$$x^{n-1}=uiff (n-1)x^{n-2},mathrm dx=,mathrm du$$
$$begin{align}
I_n
&= -x^{n-1}cdotsqrt{9-x^2}Bigg|_0^1+(n-1)int_0^1x^{n-2}sqrt{9-x^2},mathrm dxtag{1}\
&= -2sqrt2+(n-1)int_0^1frac{x^{n-2}(9-x^2)}{sqrt{9-x^2}},mathrm dxtag{2}\
&= -2sqrt2+9(n-1)int_0^1frac{x^{n-2}}{sqrt{9-x^2}},mathrm dx-(n-1)int_0^1frac{x^{n}}{sqrt{9-x^2}},mathrm dxtag{3}\
&= -2sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}tag{4}\
end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$
edited Jul 9 '18 at 17:51
tatan
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
$endgroup$
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
add a comment |
$begingroup$
$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
Using Integration By Parts
$$frac{x}{sqrt{9 - x^2}},mathrm dx=,mathrm dviff-sqrt{9-x^2}=v$$
$$x^{n-1}=uiff (n-1)x^{n-2},mathrm dx=,mathrm du$$
$$begin{align}
I_n
&= -x^{n-1}cdotsqrt{9-x^2}Bigg|_0^1+(n-1)int_0^1x^{n-2}sqrt{9-x^2},mathrm dxtag{1}\
&= -2sqrt2+(n-1)int_0^1frac{x^{n-2}(9-x^2)}{sqrt{9-x^2}},mathrm dxtag{2}\
&= -2sqrt2+9(n-1)int_0^1frac{x^{n-2}}{sqrt{9-x^2}},mathrm dx-(n-1)int_0^1frac{x^{n}}{sqrt{9-x^2}},mathrm dxtag{3}\
&= -2sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}tag{4}\
end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$
edited Jul 9 '18 at 17:51
tatan
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
$endgroup$
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
add a comment |
$begingroup$
$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
Using Integration By Parts
$$frac{x}{sqrt{9 - x^2}},mathrm dx=,mathrm dviff-sqrt{9-x^2}=v$$
$$x^{n-1}=uiff (n-1)x^{n-2},mathrm dx=,mathrm du$$
$$begin{align}
I_n
&= -x^{n-1}cdotsqrt{9-x^2}Bigg|_0^1+(n-1)int_0^1x^{n-2}sqrt{9-x^2},mathrm dxtag{1}\
&= -2sqrt2+(n-1)int_0^1frac{x^{n-2}(9-x^2)}{sqrt{9-x^2}},mathrm dxtag{2}\
&= -2sqrt2+9(n-1)int_0^1frac{x^{n-2}}{sqrt{9-x^2}},mathrm dx-(n-1)int_0^1frac{x^{n}}{sqrt{9-x^2}},mathrm dxtag{3}\
&= -2sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}tag{4}\
end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$
edited Jul 9 '18 at 17:51
tatan
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
$endgroup$
$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
Using Integration By Parts
$$frac{x}{sqrt{9 - x^2}},mathrm dx=,mathrm dviff-sqrt{9-x^2}=v$$
$$x^{n-1}=uiff (n-1)x^{n-2},mathrm dx=,mathrm du$$
$$begin{align}
I_n
&= -x^{n-1}cdotsqrt{9-x^2}Bigg|_0^1+(n-1)int_0^1x^{n-2}sqrt{9-x^2},mathrm dxtag{1}\
&= -2sqrt2+(n-1)int_0^1frac{x^{n-2}(9-x^2)}{sqrt{9-x^2}},mathrm dxtag{2}\
&= -2sqrt2+9(n-1)int_0^1frac{x^{n-2}}{sqrt{9-x^2}},mathrm dx-(n-1)int_0^1frac{x^{n}}{sqrt{9-x^2}},mathrm dxtag{3}\
&= -2sqrt2+9(n-1)I_{n-2}-(n-1)I_{n}tag{4}\
end{align}$$
After rearrangement we get
$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$
edited Jul 9 '18 at 17:51
tatan
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
edited Jul 9 '18 at 17:51
tatan
5,80262760
edited Jul 9 '18 at 17:51
tatan
5,80262760
edited Jul 9 '18 at 17:51
tatan
5,80262760
5,80262760
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
answered Dec 14 '14 at 3:22
IuʇǝƃɹɐʇoɹIuʇǝƃɹɐʇoɹ
8,35723550
8,35723550
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
add a comment |
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
$begingroup$
Thank you so much :) :) :) Integrator
$endgroup$
– jane brown
Dec 14 '14 at 4:19
add a comment |
$begingroup$
Hint:-
$x=3sin theta implies dx=3costheta dtheta$
$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
$endgroup$
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
add a comment |
$begingroup$
Hint:-
$x=3sin theta implies dx=3costheta dtheta$
$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
$endgroup$
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
add a comment |
$begingroup$
Hint:-
$x=3sin theta implies dx=3costheta dtheta$
$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
$endgroup$
Hint:-
$x=3sin theta implies dx=3costheta dtheta$
$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
answered Dec 14 '14 at 4:44
user 170039user 170039
10.5k42466
10.5k42466
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
add a comment |
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
$begingroup$
nice substitution but what happened to the limits?
$endgroup$
– abel
Dec 14 '14 at 14:35
add a comment |
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$begingroup$
Well, from the last bit, $$I_n=frac{9(n-1)nI_{n-2}-2 sqrt{2}}{n}$$ and so you can find $I_2$.
$endgroup$
– HDE 226868
Dec 14 '14 at 0:46
$begingroup$
Yeah, i figured that out as well, but how do you use integration by parts to show the required answer...
$endgroup$
– jane brown
Dec 14 '14 at 0:51