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

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