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Question Number 192687 by Mastermind last updated on 24/May/23

(1/π)∫_0 ^(2π) xcos(nx)dx Help!

$$\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{xcos}\left(\mathrm{nx}\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Answered by Subhi last updated on 24/May/23

x=u ⇛ du = dx cos(nx)dx = dv v=(1/n)∫n.cos(nx) = (1/n).sin(nx) ∫u.dv = u.v−∫v.du = x.((sin(nx))/n)−(1/n)∫sin(nx).dx x.((sin(nx))/n)+(1/n^2 )∫−n.sin(nx).dx x.((sin(nx))/n)+(1/n^2 ).cos(nx)+c (1/π)∫x.cos(nx)dx = x.((sin(nx))/(nπ))+(1/(n^2 .π))cos(nx)+c

$$ \\ $$$${x}={u}\:\:\:\:\:\Rrightarrow\:\:\:{du}\:=\:{dx} \\ $$$${cos}\left({nx}\right){dx}\:=\:{dv} \\ $$$${v}=\frac{\mathrm{1}}{{n}}\int{n}.{cos}\left({nx}\right)\:=\:\frac{\mathrm{1}}{{n}}.{sin}\left({nx}\right) \\ $$$$\int{u}.{dv}\:=\:{u}.{v}−\int{v}.{du} \\ $$$$=\:{x}.\frac{{sin}\left({nx}\right)}{{n}}−\frac{\mathrm{1}}{{n}}\int{sin}\left({nx}\right).{dx} \\ $$$${x}.\frac{{sin}\left({nx}\right)}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\int−{n}.{sin}\left({nx}\right).{dx} \\ $$$${x}.\frac{{sin}\left({nx}\right)}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }.{cos}\left({nx}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\pi}\int{x}.{cos}\left({nx}\right){dx}\:=\:{x}.\frac{{sin}\left({nx}\right)}{{n}\pi}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} .\pi}{cos}\left({nx}\right)+{c} \\ $$

Commented by Subhi last updated on 24/May/23

to find the definite integral, it will differ if (n) is integer or fraction

$${to}\:{find}\:{the}\:{definite}\:{integral},\:{it}\:{will}\:{differ}\:{if}\:\left({n}\right)\:{is}\:{integer}\:{or}\:{fraction} \\ $$$$ \\ $$

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