Two Integration things I don't understand (1 Viewer)

[RideTheLightnin]

hey guys im in cambridge, and theres two bits of working out i dont understand in the examples
On page 168, halfway down the page,
*the 'n's are down the bottom of the letter, they arent logs (as you probably already know but anyway)
In=(n-1)(I(n-2)-In)
and I dont get how this can be rearranged into
In = (n-1).I(n-2)/n

and

page 167....how the second part of the first line goes to the second....its way too tedious to type out but any help would BE GREATLY APPRECIATED!!!
cheers

 

[Riviet]

I_n=(n-1)(I_n-2-I_n)
I_n=nI_n-2-nI_n-I_n-2+I_n
nI_n=nI_n-2-I_n-2
I_n=[(n-1)I_n]/n

P.S I can't help you on your other question as I don't have my cambridge book with me now.

 

[sasquatch]

The second part:

(2/3)int(0,1) nx^n-1(1-x)sqrt(1-x)dx
= (2n/3)int(0,1) {x^n-1sqrt(1-x)}(1-x)dx
= (2n/3)int(0,1) [ {x^n-1sqrt(1-x)} - x{x^n-1sqrt(1-x)}]dx
= (2n/3)int(0,1) [ {x^n-1sqrt(1-x)} - {xⁿsqrt(1-x)}]dx

? Is that what you meant?