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)))sin ((a/(2b)))+Q((a/(2b)))cos ((a/(2b)))]


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

$I_{n} = \int_{- 1}^{1} {(1 - x^{2})}^{n} \cos (\frac{a}{2 b} x) d x$ $t o i n t e g r a t i n g b y p i e c e f o r n ⩾ 2$ $p r o v e n$ $\frac{a^{2}}{4 b^{2}} I_{n} = 2 n (2 n - 1) I_{n - 1} - 4 (n - 1) I_{n - 2}$ $p r o v e n b y r e a r r i n g t h a t$ ${(\frac{a}{2 b})}^{2 n + 1} I_{n} = n! [p (\frac{q}{2 b}) \sin (\frac{a}{2 b}) + Q (\frac{a}{2 b}) \cos (\frac{a}{2 b})]$
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