[rainbowsss]

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 I_n = 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 I_n / I_n-1

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

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

My working:
I_n = 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 I_n. It is almost I_n-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

[Dreamweaver]

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.

[DFranklin]

For the 2nd one, rewrite (1+x^3)^8 as (1+x^3)^7 + x^3 (1+x^3)^7

[rainbowsss]

(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 I_n etc, however I'm confused as to how to approach the common denominator factorisation , I assume this will also simply the extra I_n 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 .

[DFranklin]

Add (n-2)/24 I_n to both sides.