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Question Number 41913 by math khazana by abdo last updated on 15/Aug/18
letf(a)=∫0πx1+acosxdx1)findf(a)2)calculate∫0πx1+2cosxdxand∫0πx1−2cosxdx3)calculate∫0πxcosx(1+acosx)2dx4)findthevalueof∫0πxcosx(1+2cosx)2dx.
Answered by maxmathsup by imad last updated on 16/Aug/18
1)wehavef(a)=∫0πx1+a2tan(x2)1+tan2(x2)dxchangementtan(x2)=tgivef(a)=∫0∞2arctan(t)1+2at1+t22dt1+t2=∫0∞4arctan(t)1+t2+2atdt=∫0∞4arctan(t)t2+2at+1dt=4∫0∞arctan(t)t2+2at+1dt=w(1)withw(α)=∫0∞arctan(αt)t2+2at+1dtwehavew′(α)=∫0∞t(1+α2t2)(t2+2at+1)dt=αt=u∫0∞uα(1+u2)(u2α2+2auα+1)duα=∫0∞uduα2(1+u2)(u2+2αau+α2α2)=∫0∞udu(u2+1)(u2+2αau+α2)letdecomposeF(u)=u(u2+1)(u2+2αau+α2)(lettakeα>0)rootsofu2+2αau+α2Δ′=α2a2−α2=α2(a2−1)case1)∣a∣>1⇒u1=−αa+∣α∣a2−1andu2=−αa−∣α∣a2−1F(u)=au−u1+bu−u2+cu+du2+1=u(u−u1)(u−u2)(u2+1)a=u1(u1−u2)(u12+1)=∣α∣a2−1−αa2∣α∣a2−1((∣α∣a2−1−αa)2+1)=α(a2−1−a)2αa2−1(α2(a2−1−a)2+1)=a2−1−a2a2−1{α2(a2−1−a)2+1}b=u2(u2−u1)(u22+1)=−α(a+a2−1)−2αa2−1{α2(a+a2−1)2+1}limu→+∞uF(u)=0=a+b+c⇒c=−a−b⇒F(u)=au−u1+bu−u2+(−a−b)u+du2+1F(0)=0=−au1−bu2+d⇒d=au1+bu2⇒F(u)=au−u1+bu−u2+(−a−b)u+au1+bu2u2+1⇒∫0∞F(u)du=[aln∣u−u1∣+bln∣u−u2∣]0+∞−[a+b2ln(u2+1)]0+∞+(au1+bu2)[arctan(u)]0+∞=[ln(∣u−u1∣a∣u−u2∣b)−ln((u2+1)a+b2)]0+∞+π2{au1+bu2}...becontinued...
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