Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 114467 by Eric002 last updated on 19/Sep/20

∫x sin^n (x) dx

$$\int{x}\:{sin}^{{n}} \left({x}\right)\:{dx} \\ $$

Answered by Olaf last updated on 19/Sep/20

I_n (x) = ∫xsin^n xdx I_n (x)−I_(n+2) (x) = ∫xsin^n x(1−sin^2 x)dx I_n (x)−I_(n+2) (x) = ∫(xcosx)(cosxsin^n x)dx u = xcosx, u′ = cosx−xsinx v′ = cosxsin^n x, v = ((sin^(n+1) x)/(n+1)) I_n (x)−I_(n+2) (x) = (1/(n+1))xcosxsin^(n+1) x −∫(cosx−xsinx)((sin^(n+1) x)/(n+1))dx I_n (x)−I_(n+2) (x) = (1/(n+1))xcosxsin^(n+1) x −(1/(n+1))∫cosxsin^(n+1) xdx+(1/(n+1))∫xsin^(n+2) xdx I_n (x)−I_(n+2) (x) = (1/(n+1))xcosxsin^(n+1) x −(1/((n+1)(n+2)))sin^(n+2) x+(1/(n+1))I_(n+2) (x) ((n+2)/(n+1))I_(n+2) (x)−I_n (x) = ((sin^(n+1) x)/(n+1))(((sinx)/(n+2))−xcosx) I_(n+2) (x) = ((n+1)/(n+2))I_n (x)+ ((sin^(n+1) x)/(n+2))(((sinx)/(n+2))−xcosx) and I_0 (x) = ∫xdx = (x^2 /2) I_1 (x) = ∫xsinxdx = sinx−xcosx work in progress...

$$\mathrm{I}_{{n}} \left({x}\right)\:=\:\int{x}\mathrm{sin}^{{n}} {xdx} \\ $$$$\mathrm{I}_{{n}} \left({x}\right)−\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\int{x}\mathrm{sin}^{{n}} {x}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right){dx} \\ $$$$\mathrm{I}_{{n}} \left({x}\right)−\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\int\left({x}\mathrm{cos}{x}\right)\left(\mathrm{cos}{x}\mathrm{sin}^{{n}} {x}\right){dx} \\ $$$${u}\:=\:{x}\mathrm{cos}{x},\:{u}'\:=\:\mathrm{cos}{x}−{x}\mathrm{sin}{x} \\ $$$${v}'\:=\:\mathrm{cos}{x}\mathrm{sin}^{{n}} {x},\:{v}\:=\:\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}} \\ $$$$\mathrm{I}_{{n}} \left({x}\right)−\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}\mathrm{cos}{x}\mathrm{sin}^{{n}+\mathrm{1}} {x} \\ $$$$−\int\left(\mathrm{cos}{x}−{x}\mathrm{sin}{x}\right)\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}{dx} \\ $$$$\mathrm{I}_{{n}} \left({x}\right)−\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}\mathrm{cos}{x}\mathrm{sin}^{{n}+\mathrm{1}} {x} \\ $$$$−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int\mathrm{cos}{x}\mathrm{sin}^{{n}+\mathrm{1}} {xdx}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\int{x}\mathrm{sin}^{{n}+\mathrm{2}} {xdx} \\ $$$$\mathrm{I}_{{n}} \left({x}\right)−\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}\mathrm{cos}{x}\mathrm{sin}^{{n}+\mathrm{1}} {x} \\ $$$$−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\mathrm{sin}^{{n}+\mathrm{2}} {x}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$$$\frac{{n}+\mathrm{2}}{{n}+\mathrm{1}}\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)−\mathrm{I}_{{n}} \left({x}\right)\:=\:\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}\left(\frac{\mathrm{sin}{x}}{{n}+\mathrm{2}}−{x}\mathrm{cos}{x}\right) \\ $$$$\mathrm{I}_{{n}+\mathrm{2}} \left({x}\right)\:=\:\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\mathrm{I}_{{n}} \left({x}\right)+\:\frac{\mathrm{sin}^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{2}}\left(\frac{\mathrm{sin}{x}}{{n}+\mathrm{2}}−{x}\mathrm{cos}{x}\right) \\ $$$$\mathrm{and}\:\mathrm{I}_{\mathrm{0}} \left({x}\right)\:=\:\int{xdx}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{I}_{\mathrm{1}} \left({x}\right)\:=\:\int{x}\mathrm{sin}{xdx}\:=\:\mathrm{sin}{x}−{x}\mathrm{cos}{x} \\ $$$$\mathrm{work}\:\mathrm{in}\:\mathrm{progress}... \\ $$

Commented by Eric002 last updated on 19/Sep/20

well done

$${well}\:{done} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com