I-n-1-1-1-x-2-n-cos-a-2b-x-dx-to-integrating-by-piece-for-n-2-proven-a-2-4b-2-I-n-2n-2n-1-I-n-1-4-n-1-I-n-2-proven-by-rearring-that-a-2b-2n-1-I-n-n-p-q-2b-

Question Number 158858 by LEKOUMA last updated on 09/Nov/21

I_{n} = \int_{- 1}^{1} {(1 - x^{2})}^{n} \cos (\frac{a}{2 b} x) d x

t o i n t e g r a t i n g b y p i e c e f o r n ⩾ 2

p r o v e n

\frac{a^{2}}{4 b^{2}} I_{n} = 2 n (2 n - 1) I_{n - 1} - 4 (n - 1) I_{n - 2}

p r o v e n b y r e a r r i n g t h a t

{(\frac{a}{2 b})}^{2 n + 1} I_{n} = n! [p (\frac{q}{2 b}) \sin (\frac{a}{2 b}) + Q (\frac{a}{2 b}) \cos (\frac{a}{2 b})]