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

∫x sin^n (x) 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...

Commented byEric002 last updated on 19/Sep/20

well done