recurrence (1 Viewer)

[onebytwo]

given In = (from 0 to pi/2) cos^n(x)dx prove that In = ((n-1)/n)In-2

i know we have to use IBP, but i cant seem to get the In-2
any help thanks

 

[Trev]

Use partial fractions with ∫ cosx.cos^n-1x dx

 

[Riviet]

given In = (from 0 to pi/2) cos^n(x)dx prove that In = ((n-1)/n)In-2

i know we have to use IBP, but i cant seem to get the In-2
any help thanks

Use IBP with u=cos^n-1x and v'=cosx. Your uv will equal zero and with the other bit, change sin²x into 1-cos²x. Then expand, and substitute in all your I_n's and I_n-2's. Expand again and collect all the I_n terms on one side and hence isolate I_n to get what we're required to prove. Hope that helps.

Use partial fractions with ∫ cosx.cos^n-1x dx

I think you meant by parts.

 

[onebytwo]

i see now, thanks a bunch

 

[Trev]

I think you meant by parts.

Eh, same thing.

 

[Rax]

Eh, same thing.

lol Yeah, more than once I have said partial fractions when meaning Integration by parts. Too many p's.

GG