Question Number 101330 by 175 last updated on 01/Jul/20

Evaluate. ∫_(−π) ^π x^9 cos x dx

Commented bymr W last updated on 01/Jul/20

this is an odd function, since f(−x)=−f(x) for odd function: ∫_(−a) ^a f(x)dx=0 ⇒∫_(−π) ^π x^9 cos x dx=0

Commented byajfour last updated on 02/Jul/20

well what if I= ∫_0 ^( π) x^9 cos x ?

Commented bymr W last updated on 02/Jul/20

I_(2n+1) =−(2n+1)[π^(2n) +(2n)I_(2n−1) ] ....

Commented byajfour last updated on 02/Jul/20

thanks Sir!

Commented by1549442205 last updated on 02/Jul/20

F_n =∫x^n dsinx=x^n sinx−∫nx^(n−1) sinxdx =x^n sinx+n∫x^(n−1) dcosx=x^n sinx+nx^(n−1) cosx −n(n−1)∫x^(n−2) cosx=x^n sinx+nx^(n−1) cosx−n(n−1)F_(n−2) Hence,I_n =∫_0 ^π x^n cosxdx=x^n sinx∣^π _0 +nx^(n−1) cosx∣_0 ^π −nI_(n−1) I_n =−n𝛑^(n−1) −n(n−1)I_(n−2)