Question Number 75661 by peter frank last updated on 15/Dec/19
Answered by Kunal12588 last updated on 15/Dec/19
![U_n =∫x^n cos x dx u = x^n & dv = cos x dx ⇒du = nx^(n−1) & v = sin x ∴ U_n =x^n sin x − n∫x^(n−1) sin x dx ⇒U_n =x^n sin x − nV_(n−1) V_n =∫x^n sin x dx u = x^n & dv = sin x dx ⇒du = nx^(n−1) & v = −cos x dx ∴ V_n =− x^n cos x + n∫x^(n−1) cos x dx ⇒V_n = − x^n cos x + nU_(n−1) using red lines ∴ U_n = x^n sin x − n[−x^(n−1) cos x+(n−1)U_(n−2) ] ⇒ U_n = x^n sin x + nx^(n−1) cos x−n(n−1)U_(n−2) ⇒ U_n = x^(n−1) (xsin x + ncos x)−n(n−1)U_(n−2) these are just simple reduction formulae](https://www.tinkutara.com/question/Q75663.png)
$${U}_{{n}} =\int{x}^{{n}} \:{cos}\:{x}\:{dx} \\ $$$${u}\:=\:{x}^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\&\:\:\:\:{dv}\:=\:{cos}\:{x}\:{dx} \\ $$$$\Rightarrow{du}\:=\:{nx}^{{n}−\mathrm{1}} \:\:\:\&\:\:\:\:\:\:\:{v}\:=\:{sin}\:{x} \\ $$$$\therefore\:\:{U}_{{n}} ={x}^{{n}} \:{sin}\:{x}\:−\:{n}\int{x}^{{n}−\mathrm{1}} {sin}\:{x}\:{dx} \\ $$$$\Rightarrow{U}_{{n}} ={x}^{{n}} \:{sin}\:{x}\:−\:{nV}_{{n}−\mathrm{1}} \\ $$$${V}_{{n}} =\int{x}^{{n}} \:{sin}\:{x}\:{dx} \\ $$$${u}\:=\:{x}^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\&\:\:\:{dv}\:=\:{sin}\:{x}\:{dx} \\ $$$$\Rightarrow{du}\:=\:{nx}^{{n}−\mathrm{1}} \:\:\:\&\:\:\:\:\:\:{v}\:=\:−{cos}\:{x}\:{dx} \\ $$$$\therefore\:\:{V}_{{n}} \:=−\:{x}^{{n}} \:{cos}\:{x}\:+\:{n}\int{x}^{{n}−\mathrm{1}} \:{cos}\:{x}\:{dx} \\ $$$$\Rightarrow{V}_{{n}} \:=\:−\:{x}^{{n}} \:{cos}\:{x}\:+\:{nU}_{{n}−\mathrm{1}} \\ $$$${using}\:{red}\:{lines} \\ $$$$\therefore\:{U}_{{n}} =\:{x}^{{n}} {sin}\:{x}\:−\:{n}\left[−{x}^{{n}−\mathrm{1}} {cos}\:{x}+\left({n}−\mathrm{1}\right){U}_{{n}−\mathrm{2}} \right] \\ $$$$\Rightarrow\:{U}_{{n}} =\:{x}^{{n}} {sin}\:{x}\:+\:{nx}^{{n}−\mathrm{1}} {cos}\:{x}−{n}\left({n}−\mathrm{1}\right){U}_{{n}−\mathrm{2}} \\ $$$$\Rightarrow\:{U}_{{n}} =\:{x}^{{n}−\mathrm{1}} \left({xsin}\:{x}\:+\:{ncos}\:{x}\right)−{n}\left({n}−\mathrm{1}\right){U}_{{n}−\mathrm{2}} \\ $$$${these}\:{are}\:{just}\:{simple}\:{reduction}\:{formulae} \\ $$
Commented by peter frank last updated on 15/Dec/19
$${thank}\:{you} \\ $$