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Question Number 91018 by  M±th+et+s last updated on 27/Apr/20

∫_0 ^π ((sin((21x)/2))/(sin(x/2)))dx

0πsin21x2sinx2dx

Commented by mathmax by abdo last updated on 29/Apr/20

let take atry  changement (x/2) =t give  ∫_0 ^π  ((sin(((21x)/2)))/(sin((x/2))))dx =2∫_0 ^(π/2)  ((sin(21t))/(sint))dt  let w(t) =(1/(sint))  we have w(t) =(2/(e^(it) −e^(−it) )) =(2/(z−z^(−1) ))  ( z=e^(it) )  =((2z)/(z^2 −1)) =−2z((1/(1−z^2 ))) =−2z Σ_(n=0) ^∞  z^(2n)  =−2e^(it)  Σ_(n=0) ^∞  e^(2int)  ⇒  I =−4 ∫_0 ^(π/2)  sin(21t)Σ_(n=0) ^∞  e^((2n+1)it)  dt  =−4 Σ_(n=0) ^∞  ∫_0 ^(π/2)  sin(21t)e^((2n+1)t)  dt =−4 Σ_(n=0) ^∞  A_n   A_n =∫_0 ^(π/2)  sin(21t)(cos(2n+1)t +isin(2n+1)t)dt  =∫_0 ^(π/2)  sin(21t)cos(2n+1)t +i ∫_0 ^(π/2)  sin(21t)sin(2n+1)t dt  sina cosb =cos((π/2)−a)cosb =(1/2)(cos((π/2)−a+b)+cos((π/2)−a−b))  =(1/2){sin(a−b)+sin(a+b)} ⇒  ∫_0 ^(π/2)  sin(21t)cos(2n+1)t dt =(1/2)∫_0 ^(π/2) {sin(20−2n)t +sin(2n+22)t}dt  =(1/2)[−(1/(20−2n))cos(20−2n)t+(1/(2n+22))cos(2n+22)t]_0 ^(π/2)   =(1/2){−(1/(20−2n))cos(10−n)π+(1/(2n+22))cos(n+11)π  +(1/(20−2n))−(1/(2n+22))}....be continued...

lettakeatrychangementx2=tgive0πsin(21x2)sin(x2)dx=20π2sin(21t)sintdtletw(t)=1sintwehavew(t)=2eiteit=2zz1(z=eit)=2zz21=2z(11z2)=2zn=0z2n=2eitn=0e2intI=40π2sin(21t)n=0e(2n+1)itdt=4n=00π2sin(21t)e(2n+1)tdt=4n=0AnAn=0π2sin(21t)(cos(2n+1)t+isin(2n+1)t)dt=0π2sin(21t)cos(2n+1)t+i0π2sin(21t)sin(2n+1)tdtsinacosb=cos(π2a)cosb=12(cos(π2a+b)+cos(π2ab))=12{sin(ab)+sin(a+b)}0π2sin(21t)cos(2n+1)tdt=120π2{sin(202n)t+sin(2n+22)t}dt=12[1202ncos(202n)t+12n+22cos(2n+22)t]0π2=12{1202ncos(10n)π+12n+22cos(n+11)π+1202n12n+22}....becontinued...

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