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Question Number 65768 by mathmax by abdo last updated on 03/Aug/19
find the value of  ∫_0 ^∞    (dx/(x^2 −2(cosθ)x +1))
findthevalueof0dxx22(cosθ)x+1
Commented by mathmax by abdo last updated on 04/Aug/19
let I =∫_0 ^∞    (dx/(x^2 −2xcosθ +1)) ⇒I =∫_0 ^∞   (dx/(x^2 −2xcosθ +cos^2 θ +sin^2 θ))  =∫_0 ^∞     (dx/((x−cosθ)^2  +sin^2 θ)) =_(x−cosθ =∣sinθ∣u)     ∫_(−((cosθ)/(∣sinθ∣))) ^(+∞)   ((∣sinθ∣du)/(sin^2 θ(1+u^2 )))  =(1/(∣sinθ∣))[arctanu]_(−((cosθ)/(∣sinθ∣))) ^(+∞)  =(1/(∣sinθ∣)){(π/2) +arctan(((cosθ)/(∣sinθ∣)))}  case 1  sinθ>0 ⇒I =(1/(sinθ)){(π/2) +arctan((1/(tanθ)))}  =(1/(sinθ)){ (π/2) +^− (π/2) −θ}  case 2  sinθ<0 ⇒ I =−(1/(sinθ)){(π/2) −arctan(((cosθ)/(snθ)))}  =(1/(sinθ)){−(π/2) +^− (π/2) −θ}
letI=0dxx22xcosθ+1I=0dxx22xcosθ+cos2θ+sin2θ=0dx(xcosθ)2+sin2θ=xcosθ=∣sinθucosθsinθ+sinθdusin2θ(1+u2)=1sinθ[arctanu]cosθsinθ+=1sinθ{π2+arctan(cosθsinθ)}case1sinθ>0I=1sinθ{π2+arctan(1tanθ)}=1sinθ{π2+π2θ}case2sinθ<0I=1sinθ{π2arctan(cosθsnθ)}=1sinθ{π2+π2θ}