Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 41913 by math khazana by abdo last updated on 15/Aug/18

let f(a) = ∫_0 ^π     (x/(1+acosx))dx  1) find  f(a)   2) calculate  ∫_0 ^π    (x/(1+2cosx))dx and ∫_0 ^π    (x/(1−2cosx))dx  3) calculate  ∫_0 ^π   ((xcosx)/((1+acosx)^2 ))dx  4) find the value of  ∫_0 ^π    ((xcosx)/((1+2cosx)^2 ))dx .

letf(a)=0πx1+acosxdx1)findf(a)2)calculate0πx1+2cosxdxand0πx12cosxdx3)calculate0πxcosx(1+acosx)2dx4)findthevalueof0πxcosx(1+2cosx)2dx.

Answered by maxmathsup by imad last updated on 16/Aug/18

1) we have f(a) = ∫_0 ^π    (x/(1+a((2tan((x/2)))/(1+tan^2 ((x/2)))))) dx  changement tan((x/2)) =t give  f(a) =∫_0 ^∞    ((2arctan(t))/(1+((2at)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞    ((4arctan(t))/(1+t^2  +2at))dt =∫_0 ^∞    ((4arctan(t))/(t^2  +2at +1)) dt  =4 ∫_0 ^∞      ((arctan(t))/(t^2  +2at +1))dt =w(1) with w(α) =∫_0 ^∞     ((arctan(αt))/(t^2  +2at +1)) dt we have  w^′ (α) =∫_0 ^∞        (t/((1+α^2 t^2 )(t^2  +2at +1))) dt  =_(αt =u)   ∫_0 ^∞        (u/(α(1+u^2 )((u^2 /α^2 ) +2a(u/α)+1)))(du/α)  = ∫_0 ^∞         ((udu)/(α^2 (1+u^2 )(((u^2  +2αau +α^2 )/α^2 )))) =∫_0 ^∞      ((udu)/((u^2  +1)(u^2  +2αau +α^2 )))  let decompose F(u) = (u/((u^2  +1)(u^2  +2αau +α^2 ))) (let take α>0)  roots of  u^2  +2αau +α^2   Δ^′  =α^2 a^2  −α^2  =α^2 (a^2 −1)  case1) ∣a∣>1  ⇒ u_1 =−αa +∣α∣(√(a^2  −1))   and  u_2 =−αa −∣α∣(√(a^2  −1))  F(u) =  (a/(u−u_1 )) +(b/(u−u_2 ))  +((cu +d)/(u^2  +1))  =(u/((u−u_1 )(u−u_2 )(u^2  +1)))  a =  (u_1 /((u_1 −u_2 )(u_1 ^2  +1))) =((∣α∣(√(a^2 −1))−αa)/(2∣α∣(√(a^2 −1))((∣α∣(√(a^2 −1))−αa)^2 +1))) =((α((√(a^2 −1))−a))/(2α(√(a^2 −1))(α^2 ((√(a^2 −1))−a)^2 +1)))  =(((√(a^2 −1))−a)/(2(√(a^2 −1)){α^2 ((√(a^2 −1))−a)^2  +1}))  b = (u_2 /((u_2 −u_1 )(u_2 ^2  +1))) = ((−α(a+(√(a^2 −1))))/(−2α(√(a^2 −1)){ α^2 (a+(√(a^2 −1)))^2  +1}))  lim_(u→+∞) u F(u) =0 =a+b +c ⇒ c =−a−b ⇒  F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u +d)/(u^2  +1))  F(0) =0 =−(a/u_1 ) −(b/u_2 ) +d ⇒ d =(a/u_1 ) +(b/u_2 ) ⇒  F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u  +(a/u_1 )+(b/u_2 ))/(u^2  +1)) ⇒  ∫_0 ^∞   F(u)du  =[aln∣u−u_1 ∣ +bln∣u−u_2 ∣]_0 ^(+∞)   −[((a+b)/2)ln(u^2  +1)]_0 ^(+∞)   +((a/u_1 ) +(b/u_2 ))[ arctan(u)]_0 ^(+∞)   =[ ln(∣u−u_1 ∣^a ∣u−u_2 ∣^b )−ln((u^2  +1)^((a+b)/2) )]_0 ^(+∞)  +(π/2){(a/u_1 ) +(b/u_2 )}  ...be continued...

1)wehavef(a)=0πx1+a2tan(x2)1+tan2(x2)dxchangementtan(x2)=tgivef(a)=02arctan(t)1+2at1+t22dt1+t2=04arctan(t)1+t2+2atdt=04arctan(t)t2+2at+1dt=40arctan(t)t2+2at+1dt=w(1)withw(α)=0arctan(αt)t2+2at+1dtwehavew(α)=0t(1+α2t2)(t2+2at+1)dt=αt=u0uα(1+u2)(u2α2+2auα+1)duα=0uduα2(1+u2)(u2+2αau+α2α2)=0udu(u2+1)(u2+2αau+α2)letdecomposeF(u)=u(u2+1)(u2+2αau+α2)(lettakeα>0)rootsofu2+2αau+α2Δ=α2a2α2=α2(a21)case1)a∣>1u1=αa+αa21andu2=αaαa21F(u)=auu1+buu2+cu+du2+1=u(uu1)(uu2)(u2+1)a=u1(u1u2)(u12+1)=αa21αa2αa21((αa21αa)2+1)=α(a21a)2αa21(α2(a21a)2+1)=a21a2a21{α2(a21a)2+1}b=u2(u2u1)(u22+1)=α(a+a21)2αa21{α2(a+a21)2+1}limu+uF(u)=0=a+b+cc=abF(u)=auu1+buu2+(ab)u+du2+1F(0)=0=au1bu2+dd=au1+bu2F(u)=auu1+buu2+(ab)u+au1+bu2u2+10F(u)du=[alnuu1+blnuu2]0+[a+b2ln(u2+1)]0++(au1+bu2)[arctan(u)]0+=[ln(uu1auu2b)ln((u2+1)a+b2)]0++π2{au1+bu2}...becontinued...

Terms of Service

Privacy Policy

Contact: info@tinkutara.com