Question Number 104220 by bemath last updated on 20/Jul/20

∫_0 ^π  ((x^2 cos x)/((1+sin x)^2 )) dx ?

Commented bybemath last updated on 20/Jul/20

thank you both. cooll

Answered by Ar Brandon last updated on 20/Jul/20

I=∫_0 ^π ((x^2 cosx)/((1+sinx)^2 ))dx , x=π−u  I=∫_0 ^π (((π−x)^2 cos(π−x))/((1+sin(π−x))^2 ))dx=−∫(((π−x)^2 cosx)/((1+sinx)^2 ))dx  2I=∫_0 ^π ((x^2 −(π^2 −2πx+x^2 ))/((1+sinx)^2 ))∙cosxdx=∫_0 ^π ((2πx−π^2 )/((1+sinx)^2 ))cosxdx       =∫_0 ^π {((2πxcosx)/((1+sinx)^2 ))−((π^2 cosx)/((1+sinx)^2 ))}dx=∫_0 ^π ((2πxcosx)/((1+sinx)^2 ))dx+[(π^2 /(1+sinx))]_0 ^π    u(x)=2πx⇒u′(x)=2π , v′(x)=((cosx)/((1+sinx)^2 ))⇒v(x)=((−1)/(1+sinx))  ⇒2I={((−2πx)/(1+sinx))+∫((2πdx)/(1+sinx))}_0 ^π   ∫(dx/(1+sinx))=∫((1−sinx)/(cos^2 x))dx=tanx−secx  ⇒2I=2π{((−x)/(1+sinx))+tanx−secx}_0 ^π =2π(2−π)  ⇒∫_0 ^π ((x^2 cosx)/((1+sinx)^2 ))dx=(2−π)π

Answered by john santu last updated on 20/Jul/20

by parts   { ((u=x^2 ⇒du=2x dx)),((v=∫ ((d(1+sin x))/((1+sin x)^2 )) =−(1/(1+sin x)) )) :}  I= −(x^2 /(1+sin x)) ]_0 ^π  +∫_0 ^π  ((2x)/(1+sin x)) dx  I= −π^2 +∫_0 ^π  ((2x)/(1+sin x)) dx   set J = ∫_0 ^π  ((2x)/(1+sin x)) dx  replace x by π−x   J = ∫_π ^0  ((2(π−x))/(1+sin (π−x))) (−dx)  J=∫_0 ^π  ((2π−2x)/(1+sin x)) dx , so we have  2J = ∫_0 ^π  ((2π)/(1+sin x)) dx ⇒J = ∫_0 ^π (π/(1+sin x))dx  substitute t=x−(π/2)  J=∫_0 ^(π/2) π sec ^2 ((t/2)) dt = 2π  hence we conclude that   = −π^2 +2π  (JS ⊛)

Answered by mathmax by abdo last updated on 20/Jul/20

I =∫_0 ^π  ((x^2 cosx)/((1+sinx)^2 ))dx  changement x =π−t give   I =−∫_0 ^π  (((π−t)^2 (cos(π−t)))/((1+sint)^2 ))(−dt) =−∫_0 ^π   (((π^2 −2πt +t^2 )cost)/((1+sint)^2 ))dt  =−π^2  ∫_0 ^π  ((cost)/((1+sint)^2 )) +2π ∫_0 ^π  ((tcost)/((1+sint)^2 ))dt −∫_0 ^π  ((t^2  cost)/((1+sint)^2 ))dt ⇒  2I =π^2 [(1/(1+sint))]_0 ^π  +2π ∫_0 ^π  ((tcost)/((1+sint)^2 ))dt  =0 +2π ∫_0 ^(π )  ((tcost)/((1+sint)^2 ))dx  by parts u^′  =((cost)/((1+sint)^2 )) and v =t ⇒  ∫_0 ^π  ((tcost)/((1+sint)^2 ))dt =[−(t/(1+sint))]_0 ^π  −∫_0 ^π  −(1/(1+sint))dt  =−π  +∫_0 ^π   (dt/(1+sint))  changement  tan((t/2))=u give  ∫_0 ^π  (dt/(1+sint)) =∫_0 ^∞   ((2du)/((1+u^2 )(1+((2u)/(1+u^2 ))))) =∫_0 ^∞  ((2du)/(1+u^2 +2u)) =∫_0 ^∞  ((2du)/((u+1)^2 )) =[((−2)/(u+1))]_0 ^∞ =2  ⇒2I =2π{−π +2} ⇒I =2π−π^2