Question Number 101585 by Rohit@Thakur last updated on 03/Jul/20

∫_0 ^π (1/(a^2 −2a cosx + 1))dx (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 ))

Commented byRohit@Thakur last updated on 04/Jul/20

Thank you sir

Commented bymathmax by abdo last updated on 04/Jul/20

you are welcome sir.