Question Number 158858 by LEKOUMA last updated on 09/Nov/21
![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)))]](https://www.tinkutara.com/question/Q158858.png)
$${I}_{{n}} =\int_{−\mathrm{1}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{n}} \mathrm{cos}\:\left(\frac{{a}}{\mathrm{2}{b}}{x}\right){dx} \\ $$$${to}\:{integrating}\:{by}\:{piece}\:{for}\:{n}\geqslant\mathrm{2}\: \\ $$$${proven}\: \\ $$$$\frac{{a}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} }{I}_{{n}\:} =\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right){I}_{{n}−\mathrm{1}} −\mathrm{4}\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} \\ $$$${proven}\:{by}\:{rearring}\:{that}\: \\ $$$$\left(\frac{{a}}{\mathrm{2}{b}}\right)^{\mathrm{2}{n}+\mathrm{1}} {I}_{{n}} ={n}!\left[{p}\left(\frac{{q}}{\mathrm{2}{b}}\right)\mathrm{sin}\:\left(\frac{{a}}{\mathrm{2}{b}}\right)+{Q}\left(\frac{{a}}{\mathrm{2}{b}}\right)\mathrm{cos}\:\left(\frac{{a}}{\mathrm{2}{b}}\right)\right] \\ $$