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0-pi-1-a-2-2a-cosx-1-dx-a-lt-1-is-




Question Number 101585 by Rohit@Thakur last updated on 03/Jul/20
∫_0 ^π (1/(a^2 −2a cosx + 1))dx (a<1) is
0π1a22acosx+1dx(a<1)is
Answered by mathmax by abdo last updated on 03/Jul/20
f(a) =∫_0 ^π  (dx/(a^2 −2acosx +1))  changement tan((x/2))=t give  f(a) =∫_0 ^∞      ((2dt)/((1+t^2 )(a^2 −2a×((1−t^2 )/(1+t^2 )) +1))) =∫_0 ^∞   ((2dt)/(a^2  +a^2 t^2 −2a+2at^2 +1+t^2 ))  =∫_0 ^∞    ((2dt)/((a^2  +2a+1)t^2 +a^2  −2a+1)) =(2/((a+1)^2 ))∫_0 ^∞   (dt/(t^2  +(((a−1)^2 )/((a+1)^2 ))))  =(2/((a+1)^2 ))∫_0 ^∞   (dt/(t^(2 )  +(((a−1)/(a+1)))^2 ))  =_(t =((1−a)/(1+a))u)    (2/((a+1)^2 )) ∫_0 ^∞    (1/((((1−a)/(1+a)))^2 (1+u^2 )))×((1−a)/(1+a))du  =(2/((1+a)^2 ))×(((1+a)^2 )/((1−a)^2 ))×((1−a)/(1+a)) ∫_0 ^∞   (du/(1+u^2 ))  =(2/(1−a^2 ))×(π/2) =(π/(1−a^2 )) ⇒f(a) =(π/(1−a^2 ))
f(a)=0πdxa22acosx+1changementtan(x2)=tgivef(a)=02dt(1+t2)(a22a×1t21+t2+1)=02dta2+a2t22a+2at2+1+t2=02dt(a2+2a+1)t2+a22a+1=2(a+1)20dtt2+(a1)2(a+1)2=2(a+1)20dtt2+(a1a+1)2=t=1a1+au2(a+1)201(1a1+a)2(1+u2)×1a1+adu=2(1+a)2×(1+a)2(1a)2×1a1+a0du1+u2=21a2×π2=π1a2f(a)=π1a2
Commented by Rohit@Thakur last updated on 04/Jul/20
Thank you sir
Thankyousir
Commented by mathmax by abdo last updated on 04/Jul/20
you are welcome sir.
youarewelcomesir.

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