Evaluate-pi-pi-x-9-cos-x-dx-

Question Number 101330 by 175 last updated on 01/Jul/20

E v a l u a t e .

\int_{- π}^{π} x^{9} \cos x d x

Commented by mr W last updated on 01/Jul/20

t h i s i s a n o d d f u n c t i o n, s i n c e

f (- x) = - f (x)

f o r o d d f u n c t i o n :

\int_{- a}^{a} f (x) d x = 0

\Rightarrow \int_{- π}^{π} x^{9} \cos x d x = 0

Commented by ajfour last updated on 02/Jul/20

w e l l w h a t i f I = \int_{0}^{π} x^{9} \cos x ?

Commented by mr W last updated on 02/Jul/20

I_{2 n + 1} = - (2 n + 1) [π^{2 n} + (2 n) I_{2 n - 1}]

\dots .

Commented by ajfour last updated on 02/Jul/20

t h a n k s S i r!

Commented by 1549442205 last updated on 02/Jul/20

F_{n} = \int x^{n} dsinx = x^{n} sinx - \int {nx}^{n - 1} sinxdx

= x^{n} sinx + n \int x^{n - 1} dcosx = x^{n} sinx + {nx}^{n - 1} cosx

- n (n - 1) \int x^{n - 2} cosx = x^{n} sinx + {nx}^{n - 1} cosx - n (n - 1) F_{n - 2}

Hence, I_{n} = \int_{0}^{π} x^{n} cosxdx = x^{n} sinx {\underset{0}{∣}}^{π} + {nx}^{n - 1} cosx ∣_{0}^{π} - {nI}_{n - 1}

I_{n} = - n π^{n - 1} - n (n - 1) I_{n - 2}