A reduction formula for $int_0^1 x^n/sqrt{9 - x^2},mathrm dx$












3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















3












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














3












3








3


2



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 2:36









clathratus

4,8691338




4,8691338










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















4












$begingroup$

$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
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




$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much :) :) :) Integrator
    $endgroup$
    – jane brown
    Dec 14 '14 at 4:19





















2












$begingroup$

Hint:-



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



$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    nice substitution but what happened to the limits?
    $endgroup$
    – abel
    Dec 14 '14 at 14:35











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
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




$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much :) :) :) Integrator
    $endgroup$
    – jane brown
    Dec 14 '14 at 4:19


















4












$begingroup$

$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
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




$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much :) :) :) Integrator
    $endgroup$
    – jane brown
    Dec 14 '14 at 4:19
















4












4








4





$begingroup$

$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
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




$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$







share|cite|improve this answer











$endgroup$



$$I_n = int_0^1 frac{x^n}{sqrt{9 - x^2}},mathrm dx= int_0^1 x^{n-1}cdotfrac{x}{sqrt{9 - x^2}},mathrm dx$$
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




$$(n)I_{n} =9(n-1)I_{n-2} -2sqrt2$$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jul 9 '18 at 17:51









tatan

5,79762759




5,79762759










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




















  • $begingroup$
    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













2












$begingroup$

Hint:-



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



$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    nice substitution but what happened to the limits?
    $endgroup$
    – abel
    Dec 14 '14 at 14:35
















2












$begingroup$

Hint:-



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



$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    nice substitution but what happened to the limits?
    $endgroup$
    – abel
    Dec 14 '14 at 14:35














2












2








2





$begingroup$

Hint:-



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



$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$






share|cite|improve this answer









$endgroup$



Hint:-



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



$therefore displaystyleintdfrac{x^n}{sqrt{9-x^2}}dx=displaystyleintdfrac{3^nsin^ntheta cos theta}{cos theta}dtheta=3^ndisplaystyleint{sin^ntheta } dtheta$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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