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Question Number 37922 by gunawan last updated on 19/Jun/18

$$f:\mathbb{N}\to \mathbb{R}$$$$g:\mathbb{N}\to \mathbb{R}$$$$f\left(n\right)={\int }_0^{2\pi }{x}^n\sin xdx$$$$g\left(n\right)={\int }_0^{2\pi }{x}^n\cos xdx$$$$\frac{f(n+1)-f\left(n\right)}{g(n+1)-g\left(n\right)}=?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18

$$plscheckthequestion...itriedbutresult$$$$notobtained$$

Commented by ajfour last updated on 20/Jun/18

$$yes,thesamehere.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18

$${\int }_0^{2\mathrm{\Pi }}{x}^{n}sinxdx=f\left(n\right)$$$$=\mid sinx\times \frac{x^{n+1}}{n+1}{\mid }_0^{2\mathrm{\Pi }}-{\int }_0^{2\mathrm{\Pi }}{cos}^{}x.\frac{x^{n+1}}{n+1}dx$$$$f\left(n\right)=-\frac{1}{n+1}g(n+1)......\left(1\right)$$$${\int }_0^{2\mathrm{\Pi }}{x}^{n}cosxdx=g\left(n\right)$$$$=\mid cosx\frac{x^{n+1}}{n+1}{\mid }_0^{2\mathrm{\Pi }}+{\int }_0^{2\mathrm{\Pi }}sinx\frac{x^{n+1}}{n+1}dx$$$$=\mid cosx.\frac{x^{n+1}}{n+1}{\mid }_0^{2\mathrm{\Pi }}+\frac{1}{n+1}f(n+1)$$$$=1.\frac{{\left(2\mathrm{\Pi }\right)}^{n+1}}{n+1}+\frac{1}{n+1}f(n+1).....\left(2\right)$$$$\frac{f(n+1)-f\left(n\right)}{g(n+1)-g\left(n\right)}$$$$\frac{(n+1)g\left(n\right)-{\left(2\mathrm{\Pi }\right)}^{n+1}}{}$$$$contd...wait...$$$$$$

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