could someone show me how to derive the recursive formula for

(cos x )^n?

thank you!

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- Jan 9th 2010, 09:02 PMalexandrabel90recursive formula
could someone show me how to derive the recursive formula for

(cos x )^n?

thank you! - Jan 9th 2010, 09:14 PMDrexel28
- Jan 9th 2010, 09:23 PMalexandrabel90
yes.

thats what i did. and i kept getting stuck as i was not able to convert the formula to a decreasing function. i kept getting (cos x )^ (n+1) and using 1- sin^2 x = cos^2 x doesnt help..

this is my working:

integrate cos(x) (cos x)^(n-1)

= sin(x) (cos x)^(n-1) +(n-1) integrate sin^2 x (cos x)^(n-1) dx

=sin(x) (cos x)^(n-1) +(n-1) [ - cos ^n x sin x (1/n) + integrate (cos x)^(n+1) (1/n) dx ]

the answer that i'm supoosed to get is

1/n ( cos x)^(n-1) sin x + ((n-1)/n) integrate (cos x )^(n-2) dx

but i cant seem to get that no matter how i try.. - Jan 9th 2010, 09:36 PMProve It
$\displaystyle \int{\cos^n{x}\,dx} = \int{\cos^2{x}\cos^{n - 2}{x}\,dx}$

$\displaystyle = \int{(1 - \sin^2{x})\cos^{n - 2}{x}\,dx}$

$\displaystyle = \int{\cos^{n - 2}{x} - \sin^2{x}\cos^{n - 2}{x}\,dx}$

$\displaystyle = \int{\cos^{n - 2}{x}\,dx} - \int{\sin^2{x}\cos^{n - 2}{x}\,dx}$

Use integration by parts on the second integral, letting

$\displaystyle u = \sin{x}$ and $\displaystyle dv = \sin{x}\cos^{n - 2}{x}$.