Reduction Formulae Questions (1 Viewer)

tehrobzorz · Jun 6, 2009

hey all! These are reduction questions and I just need a bit of help...Im quite unsure of how to approach this one:
$In=\int \frac{x^5}{(x^2+a^2)^n}$

in terms of In-1.

and
$In=\int t^\frac{1}{2}(t^2 + 1 )^n$
also in In-1,
I got up to:

$In=\int t^\frac{1}{2}(t^2 + 1 )^n$
$=\frac{2t^\frac{3}{2}(t^2+1)^n}{3}-n\int (t^2+1)^{n-1}.\frac{4t^\frac{5}{2}}{3}$

but im unsure of what to do now... should I take out 4/3 and IBP again? =X

=\!!

kwabon · Jun 6, 2009

tehrobzorz said:
hey all! These are reduction questions and I just need a bit of help...Im quite unsure of how to approach this one:
$In=\int \frac{x^5}{(x^2+a^2)^n}$

in terms of In-1.

and
$In=\int t^\frac{1}{2}(t^2 + 1 )^n$
also in In-1,
I got up to:

$In=\int t^\frac{1}{2}(t^2 + 1 )^n$
$=\frac{2t^\frac{3}{2}(t^2+1)^n}{3}-n\int (t^2+1)^{n-1}.\frac{4t^\frac{5}{2}}{3}$

but im unsure of what to do now... should I take out 4/3 and IBP again? =X

=\!!

could you post up the whole question, because a reduction question usually has something to prove, i just do not know where i am going without that bit....so yeh

if you want i will still post up my solution

tehrobzorz · Jun 6, 2009

it just says " find a formula finding In in terms of In-1"

and yeah, please post your solutions thanks!

Trebla · Jun 6, 2009

$I_n=\int \dfrac{x^5}{(x^2+a^2)^n}\,dx\\\\u = (x^2+a^2)^{-n},\,du=-2xn(x^2+a^2)^{-n-1}\,dx\\\\dv=x^5\,dx,\,v=\frac{x^6}{6}\\\\I_n=\dfrac{x^6}{6(x^2+a^2)^n}+\dfrac{n}{3}\int x^7(x^2+a^2)^{-n-1}\,dx\\\\=\dfrac{x^6}{6(x^2+a^2)^n}+\dfrac{n}{3}\int x^5(x^2+a^2-a^2)(x^2+a^2)^{-n-1}\,dx\\\\=\dfrac{x^6}{6(x^2+a^2)^n}+\dfrac{n}{3}\int x^5 (x^2+a^2)^{-n}-a^2x^5(x^2+a^2)^{-n-1}\,dx\\\\=\dfrac{x^6}{6(x^2+a^2)^n}+\dfrac{n}{3}I_n-\frac{na^2}{3}I_{n+1}\\\\\frac{na^2}{3}I_{n+1}=\dfrac{n-3}{3}I_n+\dfrac{x^6}{6(x^2+a^2)^n}\\\\I_{n+1}=\dfrac{n-3}{na^2}I_n+\dfrac{x^6}{2na^2(x^2+a^2)^n}$

$$Replace $n$ with $n-1$ gives$\\\\I_{n}=\dfrac{n-4}{(n-1)a^2}I_{n-1}+\dfrac{x^6}{2(n-1)a^2(x^2+a^2)^{n-1}}$

Check my working, it looks really messy...it might be simpler if it had limits or something...

tehrobzorz · Jun 6, 2009

mmk ..cheers! so on a scale of 1-10.. what difficulty was it for you? 0_0" reductions involving trig seems a whole lot easier to me X_X

I dont get steps 2-3 though.. like is that the systematic way to simplify into terms of In? =X

Dumbledore · Jun 6, 2009

$\int t^{\frac{1}{2}}(t^{2}+1)^{n-1}[t^{2}+1]dt\\\int t^{\frac{5}{2}}(t^{2}+1)^{n-1}dt + I_{n-1}\\ \frac{1}{n}\int [nt(t^{2}+1)^{n-1}].t^{\frac{3}{2}}dt + I_{n-1} \\\frac{1}{n}\int t^{\frac{3}{2}}d[(t^{2}+1)^{n}]+ I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n} - \frac{3}{2n}\int t^{\frac{1}{2}}(t^{2}+1)^{n}dt +I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n}+\frac{3}{2n}I_{n}+I_{n-1}$
check for mistakes, i have plenty of them

gurmies · Jun 7, 2009

Dumbledore said:
$\int t^{\frac{1}{2}}(t^{2}+1)^{n-1}[t^{2}+1]dt\\\int t^{\frac{5}{2}}(t^{2}+1)^{n-1}dt + I_{n-1}\\ \frac{1}{n}\int [nt(t^{2}+1)^{n-1}].t^{\frac{3}{2}}dt + I_{n-1} \\\frac{1}{n}\int t^{\frac{3}{2}}d[(t^{2}+1)^{n}]+ I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n} - \frac{3}{2n}\int t^{\frac{1}{2}}(t^{2}+1)^{n}dt +I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n}+\frac{3}{2n}I_{n}+I_{n-1}$
check for mistakes, i have plenty of them

Should be 1/2n because of the t^2 I think

Aerath · Jun 8, 2009

I think.

tehrobzorz · Jun 8, 2009

okay! thanks everyone!

=X

madsam · Jun 11, 2009

Dumbledore said:
$\int t^{\frac{1}{2}}(t^{2}+1)^{n-1}[t^{2}+1]dt\\\int t^{\frac{5}{2}}(t^{2}+1)^{n-1}dt + I_{n-1}\\ \frac{1}{n}\int [nt(t^{2}+1)^{n-1}].t^{\frac{3}{2}}dt + I_{n-1} \\\frac{1}{n}\int t^{\frac{3}{2}}d[(t^{2}+1)^{n}]+ I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n} - \frac{3}{2n}\int t^{\frac{1}{2}}(t^{2}+1)^{n}dt +I_{n-1} \\\frac{1}{n}t^{\frac{3}{2}}(t^{2}+1)^{n}+\frac{3}{2n}I_{n}+I_{n-1}$
check for mistakes, i have plenty of them

$\int t\tfrac{1}{2}\left ( t^{2}+1 \right )^{n}dt$

Using the form u' . v as partial fractions

$V= \left ( t^{2}+1 \right )\\V' = \left ( t^2 +1)^{n-1} \right )(n-1)(2t)\\u = 2t^\frac{3}{2}/3\\u' = t^\frac{1}{2}\\I_{n}= \frac{2}{3}t^\frac{3}{2}(t^2+1)^{n} - \int \frac{2}{3}t^\frac{3}{2}.2t(n-1).(t^2+1)^{n-1}\\= \frac{2}{3}t^\frac{3}{2}(t^2+1)^{n} - \frac{2}{3}(n-1)\int t^\frac{3}{2}.2t.(t^2+1)^{n-1}\\= \frac{2}{3}t^\frac{3}{2}(t^2+1)^{n} - \frac{4}{3}(n-1)I_{n-1}$

Sorry if its messy, its my first time using the latex program thingo

edit: wait... just ignore me

funkyknight · Jun 13, 2009

hey.. with the first question u said u didnt understand step 2-3

the person used integration by parts...

(f= integral sign)

fuv'= uv - fvu'

tehrobzorz · Jun 13, 2009

thanks for noticing funkynight! but, i already knew about IBP. I wasn't sure if taking away the term a^2 was a systematic approach to most reduction questions and how he made it to the power of -n.

=X

Reduction Formulae Questions (1 Viewer)

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