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# Reduction Forumla Integration.

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1. Reduction Forumla Integration.
1) Use a reduction formula to integrate:
(x^3)(sin x).

The issue I am having, I shall explain it in terms of the forumla to make it simpler:
udv/dx=uv - vdu/dx (say).

Let In = x^n (sinx)
Let u = x^n and dv/dv = sin x , -vdu/dx becomes: (n)(x^n-1) (-cos x),
And, this I am unable to express in terms of In / In-1

If anyone could shed some light that'll be awesome , thanks guys.

2) If In = (x^n)[(1+x^3)^7] dx, prove that:
In = 1/(n+22) [x^n-2((1+x^3)^8)-(n-2)]In-3

My working:
In = x^2(x^n-2)[(1+x^3)^7] dx

By parts, using the formula arranged as above and letting:
dv/dx = (x^2)((1+x^3)^7)
v = ((1+X^3)^8) / 24
u = (x^n-2)
du/dx= (n-2)(x^n-3)

In = 1/24 [(x^n-2)(1+x^3)^8] -(n-2/24) [(1+x^3)^8)(x^n-3) dx ]

Again, the last term I am no able to express in terms of In. It is almost In-3 but not quite - I have tried simplyfying and expanding the (n-2) out, but had no luck...

Again if anyone could offer any assitance, that'll be greatly appreciated, taaa
2. Re: Reduction Forumla Integration.
1) Your on the right track. You have to do integration by parts twice. This enables you to express it in terms of In/In-2.

Could you put the second one in latex? It's hard to understand and tedious to follow.
3. Re: Reduction Forumla Integration.
For the 2nd one, rewrite (1+x^3)^8 as (1+x^3)^7 + x^3 (1+x^3)^7
4. Re: Reduction Forumla Integration.
(Original post by DFranklin)
For the 2nd one, rewrite (1+x^3)^8 as (1+x^3)^7 + x^3 (1+x^3)^7
Thanks for the advice, I think it has led me in the right direction, but I'm still struggling... all is now in terms of In etc, however I'm confused as to how to approach the common denominator factorisation , I assume this will also simply the extra In term I seem to have...?

In = 1/24 [(x^n-2)(1+x^3)^8] -(n-2/24) [(1+x^3)^7)(x^n)+(1+x^3)^7)(x^n-3) dx ]
In = 1/24 [(x^n-2)(1+x^3)^8] -[(n-2/24) (In+In-3)]

Thanks alot .
Last edited by rainbowsss; 15-03-2012 at 15:23.
5. Re: Reduction Forumla Integration.
Add (n-2)/24 I_n to both sides.

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Last updated: March 15, 2012
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