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Question Number 37922 by gunawan last updated on 19/Jun/18
f : N → R g : N → R f(n)=∫_0 ^(2π) x^n sin x dx g(n)=∫_0 ^(2π) x^n cos x dx ((f(n+1)−f(n))/(g(n+1)−g(n)))=?
$${f}\::\:\mathbb{N}\:\rightarrow\:\mathbb{R} \\ $$$${g}\::\:\mathbb{N}\:\rightarrow\:\mathbb{R} \\ $$$${f}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {x}^{{n}} \mathrm{sin}\:{x}\:{dx} \\ $$$${g}\left({n}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {x}^{{n}} \mathrm{cos}\:{x}\:{dx} \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)−{f}\left({n}\right)}{{g}\left({n}+\mathrm{1}\right)−{g}\left({n}\right)}=? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18
pls check the question...i tried but result not obtained
$${pls}\:{check}\:{the}\:{question}…{i}\:{tried}\:{but}\:{result} \\ $$$${not}\:{obtained} \\ $$
Commented by ajfour last updated on 20/Jun/18
yes, the same here.
$${yes},\:{the}\:{same}\:{here}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18
∫_0 ^(2Π) x^n sinx dx=f(n) =∣sinx×(x^(n+1) /(n+1))∣_0 ^(2Π) −∫_0 ^(2Π) cos^ x.(x^(n+1) /(n+1))dx f(n)=−(1/(n+1))g(n+1)......(1) ∫_0 ^(2Π) x^n cosxdx=g(n) =∣cosx(x^(n+1) /(n+1))∣_0 ^(2Π) +∫_0 ^(2Π) sinx(x^(n+1) /(n+1))dx =∣cosx.(x^(n+1) /(n+1))∣_0 ^(2Π) +(1/(n+1))f(n+1) =1.(((2Π)^(n+1) )/(n+1))+(1/(n+1))f(n+1).....(2) ((f(n+1)−f(n))/(g(n+1)−g(n))) (((n+1)g(n)−(2Π)^(n+1) )/) contd ...wait...
$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {x}^{{n}} {sinx}\:{dx}={f}\left({n}\right) \\ $$$$=\mid{sinx}×\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{2}\Pi} −\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {cos}^{} {x}.\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{dx} \\ $$$${f}\left({n}\right)=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{g}\left({n}+\mathrm{1}\right)……\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {x}^{{n}} {cosxdx}={g}\left({n}\right) \\ $$$$=\mid{cosx}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{2}\Pi} +\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {sinx}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{dx} \\ $$$$=\mid{cosx}.\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{2}\Pi} +\frac{\mathrm{1}}{{n}+\mathrm{1}}{f}\left({n}+\mathrm{1}\right) \\ $$$$=\mathrm{1}.\frac{\left(\mathrm{2}\Pi\right)^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}{f}\left({n}+\mathrm{1}\right)…..\left(\mathrm{2}\right) \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)−{f}\left({n}\right)}{{g}\left({n}+\mathrm{1}\right)−{g}\left({n}\right)} \\ $$$$\frac{\left({n}+\mathrm{1}\right){g}\left({n}\right)−\left(\mathrm{2}\Pi\right)^{{n}+\mathrm{1}} }{} \\ $$$${contd}\:…{wait}… \\ $$$$ \\ $$

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