Question Number 115014 by mathdave last updated on 23/Sep/20

if I_n =∫_x ^(π/2) xcos^n xdx,where n≻1 show  that I_n =((n(n−1)I_(n−2) −1)/n^2 ) and then  evaluate  ∫_x ^(π/2) xcos^8 xdx

Answered by Olaf last updated on 23/Sep/20

  I_n  = ∫_0 ^(π/2) xcos^n xdx  I_n −I_(n+2)  = ∫_0 ^(π/2) xcos^n x(1−cos^2 x)dx  I_n −I_(n+2)  = −∫_0 ^(π/2) (xsinx)(−sinxcos^n x)dx  I_n −I_(n+2)  = −[xsinx((cos^(n+1) x)/(n+1))]_0 ^(π/2)   +∫_0 ^(π/2) (sinx+xcosx)((cos^(n+1) x)/(n+1))dx  I_n −I_(n+2)  = −∫_0 ^(π/2) (−sinx((cos^(n+1) x)/(n+1)))dx  +(1/(n+1))I_(n+2)   I_n −I_(n+2)  = −[((cos^(n+2) x)/((n+1)(n+2)))]_0 ^(π/2) +(1/(n+1))I_(n+2)   I_n  =  (1/((n+1)(n+2)))+I_(n+2)  +(1/(n+1))I_(n+2)   I_n  = ((n+2)/(n+1))I_(n+2) +(1/((n+1)(n+2)))  I_(n+2)  = ((n+1)/(n+2))I_n −(1/((n+2)^2 )) (1)  I_n  = ((n(n−1)I_(n−2) −1)/n^2 )  I_0  = ∫_0 ^(π/2) xdx = [(x^2 /2)]_0 ^(π/2)  = (π^2 /8)  With (1) : I_2  = (1/2)((π^2 /8))−(1/4) = (π^2 /(16))−(1/4)  I_4  = (3/4)((π^2 /(16))−(1/4))−(1/(16)) = ((3π^2 )/(64))−(1/4)  I_6  = (5/6)(((3π^2 )/(64))−(1/4))−(1/(36)) = ((5π^2 )/(128))−((17)/(72))  I_8  = (7/8)(((5π^2 )/(128))−((17)/(72)))−(1/(64)) = ((35π^2 )/(1024))−(2/9)  I_8  = ∫_0 ^(π/2) xcos^8 xdx = ((35π^2 )/(1024))−(2/9)

Commented byTawa11 last updated on 06/Sep/21

great sir