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reduction formula

Hi im stuck on this question

let I(n)=intergral between 0 and 1 of cosh^n x
by considering d/dx(sinhx *cosh^(n-1) x) show that nI(n)=ab^(n-1) +(n-1)I(n-2) where a=sinh 1 and b=cosh 1

any help much appreciated!

thanks

Reply 1

yazan_l

I_n = int coshⁿx dx
= int coshx . cosh^n-1x dx
= cosh^n-1x sinhx - (n-1) int cosh^n-2x . sinh²x dx
= cosh^n-1x sinhx - (n-1) int cosh^n-2x . (cosh²x - 1) dx
= cosh^n-1x sinhx - (n-1) int coshⁿx dx + (n-1) int cosh^n-2x dx
I_n = cosh^n-1x sinhx - (n-1)I_n + (n-1) I_n-2

I_n (n-1+1) = cosh^n-1x sinhx + (n-1) I_n-2

n I_n = cosh^n-1x sinhx + (n-1) I_n-2

Edit: forgot to plug the limits in...
n I_n = cosh^n-1x sinhx]₀¹ + (n-1) I_n-2
n I_n = cosh^n-11 sinh1 - cosh^n-10 sinh0 + (n-1) I_n-2
n I_n = cosh^n-11 sinh1 + (n-1) I_n-2

Reply 2

meef cheese

Integrate by parts with:

u = cosh^n-1(x) => u' = (n-1)cosh^n-2(x).sinh(x)

v' = cosh(x) => v = sinh(x)

I_n = uv - u'v
= {between 1 and 0} [cosh^n-1(x).sinhx] - (n-1) INT cosh^n-2(x).sinh²(x)

Then use sinh²(x) = cosh²(x) - 1
to get

I_n = {between 1 and 0} [cosh^n-1(x).sinhx] - (n-1) INT coshⁿ(x) - cosh^n-2(x)

and hopefully you can do it from there

EDIT: or just read Yazan's post :wink: