Reduction formula for $intfrac{dx}{(ax^2+b)^n}$











up vote
0
down vote

favorite












I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}\I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=sqrt{frac ba}t$, and it gave me
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intfrac{mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=tan u$:
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intcot^{n-1}u mathrm{d}u$$
I then used the $cot^nu$ reduction formula to find
$$I_n=frac{-b^{1/2-n}}{a^{1/2}}bigg(frac{cot^{n-2}u}{n-2}+intcot^{n-3}u mathrm{d}ubigg)$$
$$I_n=frac{-b^{1/2-n}cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula.



Could someone provide a derivation of the reduction formula? Thanks.






calculus integration proof-explanation indefinite-integrals reduction-formula







share|cite|improve this question
























  • I think you've found the reduction formula depends on $b$.
    – Nosrati
    Nov 15 at 4:48










  • @Nosrati how so?
    – clathratus
    Nov 15 at 4:49










  • Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
    – Travis
    Nov 15 at 20:03















up vote
0
down vote

favorite












I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}\I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=sqrt{frac ba}t$, and it gave me
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intfrac{mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=tan u$:
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intcot^{n-1}u mathrm{d}u$$
I then used the $cot^nu$ reduction formula to find
$$I_n=frac{-b^{1/2-n}}{a^{1/2}}bigg(frac{cot^{n-2}u}{n-2}+intcot^{n-3}u mathrm{d}ubigg)$$
$$I_n=frac{-b^{1/2-n}cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula.



Could someone provide a derivation of the reduction formula? Thanks.






calculus integration proof-explanation indefinite-integrals reduction-formula







share|cite|improve this question
























  • I think you've found the reduction formula depends on $b$.
    – Nosrati
    Nov 15 at 4:48










  • @Nosrati how so?
    – clathratus
    Nov 15 at 4:49










  • Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
    – Travis
    Nov 15 at 20:03













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}\I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=sqrt{frac ba}t$, and it gave me
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intfrac{mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=tan u$:
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intcot^{n-1}u mathrm{d}u$$
I then used the $cot^nu$ reduction formula to find
$$I_n=frac{-b^{1/2-n}}{a^{1/2}}bigg(frac{cot^{n-2}u}{n-2}+intcot^{n-3}u mathrm{d}ubigg)$$
$$I_n=frac{-b^{1/2-n}cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula.



Could someone provide a derivation of the reduction formula? Thanks.






calculus integration proof-explanation indefinite-integrals reduction-formula







share|cite|improve this question















I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=intfrac{mathrm{d}x}{(ax^2+b)^n}\I_n=frac{x}{2b(n-1)(ax^2+b)^{n-1}}+frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=sqrt{frac ba}t$, and it gave me
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intfrac{mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=tan u$:
$$I_n=frac{b^{1/2-n}}{a^{1/2}}intcot^{n-1}u mathrm{d}u$$
I then used the $cot^nu$ reduction formula to find
$$I_n=frac{-b^{1/2-n}}{a^{1/2}}bigg(frac{cot^{n-2}u}{n-2}+intcot^{n-3}u mathrm{d}ubigg)$$
$$I_n=frac{-b^{1/2-n}cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula.



Could someone provide a derivation of the reduction formula? Thanks.






calculus integration proof-explanation indefinite-integrals reduction-formula




calculus integration proof-explanation indefinite-integrals reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 0:56

























asked Nov 15 at 4:41









clathratus

2,544325




2,544325












  • I think you've found the reduction formula depends on $b$.
    – Nosrati
    Nov 15 at 4:48










  • @Nosrati how so?
    – clathratus
    Nov 15 at 4:49










  • Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
    – Travis
    Nov 15 at 20:03


















  • I think you've found the reduction formula depends on $b$.
    – Nosrati
    Nov 15 at 4:48










  • @Nosrati how so?
    – clathratus
    Nov 15 at 4:49










  • Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
    – Travis
    Nov 15 at 20:03
















I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 at 4:48




I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 at 4:48












@Nosrati how so?
– clathratus
Nov 15 at 4:49




@Nosrati how so?
– clathratus
Nov 15 at 4:49












Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
– Travis
Nov 15 at 20:03




Note that this method only works (at least without introducing complex numbers, which requires some care to resolve) if $a > 0, b geq 0$.
– Travis
Nov 15 at 20:03










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.






share|cite|improve this answer



















  • 1




    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
    – clathratus
    Nov 15 at 5:01










  • I'm glad you found it useful, cheers!
    – Travis
    Nov 15 at 5:05










  • Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
    – clathratus
    Nov 15 at 18:52










  • I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
    – Travis
    Nov 15 at 19:40












  • Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
    – clathratus
    Nov 15 at 19:41











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999219%2freduction-formula-for-int-fracdxax2bn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.






share|cite|improve this answer



















  • 1




    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
    – clathratus
    Nov 15 at 5:01










  • I'm glad you found it useful, cheers!
    – Travis
    Nov 15 at 5:05










  • Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
    – clathratus
    Nov 15 at 18:52










  • I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
    – Travis
    Nov 15 at 19:40












  • Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
    – clathratus
    Nov 15 at 19:41















up vote
1
down vote



accepted










Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.






share|cite|improve this answer



















  • 1




    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
    – clathratus
    Nov 15 at 5:01










  • I'm glad you found it useful, cheers!
    – Travis
    Nov 15 at 5:05










  • Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
    – clathratus
    Nov 15 at 18:52










  • I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
    – Travis
    Nov 15 at 19:40












  • Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
    – clathratus
    Nov 15 at 19:41













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.






share|cite|improve this answer














Hint The appearance of the term in $frac{x}{(a x^2 + b)^{n - 1}}$ suggests applying integration by parts with $dv = dx$ and thus $u = (a x^2 + b)^{-n}$. Renaming $n$ to $m$ we get
$$I_m = u v - int v ,du = frac{x}{(a x^2 + b)^m} + 2 m int frac{a x^2 ,dx}{(a x^2 + b)^{m + 1}} .$$
Now, the integral on the right can be rewritten as a linear combination $p I_{m + 1} + qI_m$, so we can solve for $I_{m + 1}$ in terms of $I_m$ and replace $m$ with $n - 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 15 at 18:59

























answered Nov 15 at 5:00









Travis

59.2k766144




59.2k766144








  • 1




    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
    – clathratus
    Nov 15 at 5:01










  • I'm glad you found it useful, cheers!
    – Travis
    Nov 15 at 5:05










  • Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
    – clathratus
    Nov 15 at 18:52










  • I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
    – Travis
    Nov 15 at 19:40












  • Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
    – clathratus
    Nov 15 at 19:41














  • 1




    Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
    – clathratus
    Nov 15 at 5:01










  • I'm glad you found it useful, cheers!
    – Travis
    Nov 15 at 5:05










  • Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
    – clathratus
    Nov 15 at 18:52










  • I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
    – Travis
    Nov 15 at 19:40












  • Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
    – clathratus
    Nov 15 at 19:41








1




1




Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 at 5:01




Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 at 5:01












I'm glad you found it useful, cheers!
– Travis
Nov 15 at 5:05




I'm glad you found it useful, cheers!
– Travis
Nov 15 at 5:05












Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
– clathratus
Nov 15 at 18:52




Based on what you said I was able to prove it last night. That's when I learned just how damn elegant it was! I would upvote this answer twice if I could. Where did you learn such a simple trick?
– clathratus
Nov 15 at 18:52












I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
– Travis
Nov 15 at 19:40






I don't think I learned it anywhere in particular---really, the only trick here is reindexing from $n$ to $m = n - 1$. At any rate, I'm happy you found the method illuminating!
– Travis
Nov 15 at 19:40














Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
– clathratus
Nov 15 at 19:41




Yeah it was just so simple yet so unexpected. I'm always used to proving reduction formulas without having to use that trick.
– clathratus
Nov 15 at 19:41


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999219%2freduction-formula-for-int-fracdxax2bn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more