Mixing
Reduction formula for $intfrac{dx}{(ax^2+b)^n}$
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
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
add a comment |
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
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
@Nosrati how so?
– clathratus
Nov 15 '18 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 '18 at 20:03
add a comment |
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
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
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
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
edited Nov 23 '18 at 0:56
clathratus
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
asked Nov 15 '18 at 4:41
clathratusclathratus
3,362331
3,362331
I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
@Nosrati how so?
– clathratus
Nov 15 '18 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 '18 at 20:03
add a comment |
I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
@Nosrati how so?
– clathratus
Nov 15 '18 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 '18 at 20:03
I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
@Nosrati how so?
– clathratus
Nov 15 '18 at 4:49
@Nosrati how so?
– clathratus
Nov 15 '18 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 '18 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 '18 at 20:03
add a comment |
1 Answer
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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 $n - 1$.
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 at 19:41
add a comment |
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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 $n - 1$.
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 at 19:41
add a comment |
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 $n - 1$.
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 at 19:41
add a comment |
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 $n - 1$.
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
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 $n - 1$.
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
edited Nov 15 '18 at 18:59
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
answered Nov 15 '18 at 5:00
TravisTravis
59.9k767146
59.9k767146
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 at 19:41
add a comment |
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 at 19:41
1
1
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh. Thanks (+1)
– clathratus
Nov 15 '18 at 5:01
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 at 5:05
I'm glad you found it useful, cheers!
– Travis
Nov 15 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 19:41
add a comment |
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I think you've found the reduction formula depends on $b$.
– Nosrati
Nov 15 '18 at 4:48
@Nosrati how so?
– clathratus
Nov 15 '18 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 '18 at 20:03