x-n-cos-nx-dx-

Question Number 152186 by Tawa11 last updated on 26/Aug/21

$$\int\mathrm{x}^{\mathrm{n}} \:\mathrm{cos}\left(\mathrm{nx}\right)\:\mathrm{dx} \\ $$

Answered by mindispower last updated on 26/Aug/21

nx=y ⇔(1/n^(n+1) )∫y^n cos(x)dx =(1/(2.n^(n+1) ))(∫y^n e^(iy) dy+∫y^n e^(−iy) dy) iy=−t,in first −iy=−w 2nd⇒ =(1/2).(1/n^(n+1) )(i^(n+1) ∫t^n e^(−t) dt+(−i)^(n+1) ∫w^n e^(−w) dt) (i^(n+1) /(2.n^(n+1) ))∫t^n e^(−t) dt+(((−i)^(n+1) )/n^(n+1) )∫w^n e^(−w) dw recall Γ(a,x) incomplet Gamma function Γ(a,x): ∫_0 ^a t^(x−1) e^(−t) dt We Get (i^(n+1) /(2.n^(n+1) ))Γ(t,n+1)+(((−i)^(n+1) )/(2.n^(n+1) ))Γ(w,n+1) =(1/2)((i/n))^(n+1) Γ(−inx,n+1)+(1/2)(−(i/n))^(n+1) Γ(inx,n+1) ∫x^n cos(nx)dx =(1/2)((i/n))^(n+1) (Γ(−inx,n+1)+(−1)^(n+1) Γ(inx,n+1))+c

$${nx}={y} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{{n}^{{n}+\mathrm{1}} }\int{y}^{{n}} {cos}\left({x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}.{n}^{{n}+\mathrm{1}} }\left(\int{y}^{{n}} {e}^{{iy}} {dy}+\int{y}^{{n}} {e}^{−{iy}} {dy}\right) \\ $$$${iy}=−{t},{in}\:{first}\:−{iy}=−{w}\:\:\mathrm{2}{nd}\Rightarrow \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{{n}^{{n}+\mathrm{1}} }\left({i}^{{n}+\mathrm{1}} \int{t}^{{n}} {e}^{−{t}} {dt}+\left(−{i}\right)^{{n}+\mathrm{1}} \int{w}^{{n}} {e}^{−{w}} {dt}\right) \\ $$$$\frac{{i}^{{n}+\mathrm{1}} }{\mathrm{2}.{n}^{{n}+\mathrm{1}} }\int{t}^{{n}} {e}^{−{t}} {dt}+\frac{\left(−{i}\right)^{{n}+\mathrm{1}} }{{n}^{{n}+\mathrm{1}} }\int{w}^{{n}} {e}^{−{w}} {dw} \\ $$$${recall}\:\Gamma\left({a},{x}\right)\:{incomplet}\:{Gamma}\:{function} \\ $$$$\Gamma\left({a},{x}\right):\:\int_{\mathrm{0}} ^{{a}} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$${We}\:{Get}\: \\ $$$$\frac{{i}^{{n}+\mathrm{1}} }{\mathrm{2}.{n}^{{n}+\mathrm{1}} }\Gamma\left({t},{n}+\mathrm{1}\right)+\frac{\left(−{i}\right)^{{n}+\mathrm{1}} }{\mathrm{2}.{n}^{{n}+\mathrm{1}} }\Gamma\left({w},{n}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{i}}{{n}}\right)^{{n}+\mathrm{1}} \Gamma\left(−{inx},{n}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{{i}}{{n}}\right)^{{n}+\mathrm{1}} \Gamma\left({inx},{n}+\mathrm{1}\right) \\ $$$$\int{x}^{{n}} {cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{i}}{{n}}\right)^{{n}+\mathrm{1}} \left(\Gamma\left(−{inx},{n}+\mathrm{1}\right)+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \Gamma\left({inx},{n}+\mathrm{1}\right)\right)+{c} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 26/Aug/21

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by mindispower last updated on 28/Aug/21

$${pleasur}\:{sir} \\ $$$$ \\ $$

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