Reduction formula >.< (1 Viewer)

[stuka]

Hmmm trying to do the 2008 hsc reduction question question 4 ish i think, thought it looked ok but im having a bit of trouble getting my head round part of it. The coroneos solutions are extremely unhelpful .

question is, let In= integral from 0-pi/4 of tan^(2n) theta,
thus prove In= 1/(2n-1) - (In-1)
where the In is the integration symbol thing.
I started off with splitting it into tan^(2n-2) and tan^2 then going from there, but i keep on getting I(n-2) instead of I(n-1)....

i hate 4 unit.

 

[MOP777]

http://www.hsccoaching.com/Resources/2008Ext2.pdf

question 3c?

The way he did it was by working out In + I(n-1) on the left side then subrating I(n-1) at the end. Quite smart actually

 

[Trebla]

Hmmm trying to do the 2008 hsc reduction question question 4 ish i think, thought it looked ok but im having a bit of trouble getting my head round part of it. The coroneos solutions are extremely unhelpful .

question is, let In= integral from 0-pi/4 of tan^(2n) theta,
thus prove In= 1/(2n-1) - (In-1)
where the In is the integration symbol thing.
I started off with splitting it into tan^(2n-2) and tan^2 then going from there, but i keep on getting I(n-2) instead of I(n-1)....

i hate 4 unit.

What you've done is fine. Note that you've actually got I(n-1) due to the tan^(2(n-1)) theta

 

[stuka]

Ah sweet cheers guys, haha yeah i know coroneos's way is smart, just is wasnt really useful in terms of trying to get it algebraically, the long way . Thanks again.