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I-0-pi-2-2304cosx-cos4x-8cos2x-15-2-dx-

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Question Number 129710 by SOMEDAVONG last updated on 18/Jan/21
I=∫_0 ^(π/2) ((2304cosx)/((cos4x−8cos2x+15)^2 ))dx
I=∫0π22304cosx(cos4x−8cos2x+15)2dx
Answered by MJS_new last updated on 18/Jan/21
2304∫((cos x)/((cos 4x −8cos 2x +15)^2 ))dx= [t=sin x → dx=(dt/(cos x))] =36∫(dt/((t^2 −t+1)^2 (t^2 +t+1)^2 ))= [Ostrogradski′s Method] =−((6t(t−1)(t+1))/((t^2 −t+1)(t^2 +t+1)))−6∫((t^2 −5)/((t^2 −t+1)(t^2 +t+1)))dt −6∫((t^2 −5)/((t^2 −t+1)(t^2 +t+1)))dt= =−3∫((6t−5)/(t^2 −t+1))dt+3∫((6t+5)/(t^2 +t+1))dt= =6∫(dt/(t^2 −t+1))−9∫((2t−1)/(t^2 −t+1))dt+6∫(dt/(t^2 +t+1))+9∫((2t+1)/(t^2 +t+1))dt= =4(√3)arctan (((√3)(2t−1))/3) −9ln (t^2 −t+1) +4(√3)arctan (((√3)(2t+1))/3) +9ln (t^2 +t+1) ⇒ we have 2304∫_0 ^(π/2) ((cos x)/((cos 4x −8cos 2x +15)^2 ))dx= [−((6t(t−1)(t+1))/((t^2 −t+1)(t^2 +t+1)))+4(√3)(arctan (((√3)(2t−1))/3) +arctan (((√3)(2t+1))/3))+9ln ((t^2 +t+1)/(t^2 −t+1))]_0 ^1 = =2π(√3)+9ln 3
2304∫cosx(cos4x−8cos2x+15)2dx=[t=sinx→dx=dtcosx]=36∫dt(t2−t+1)2(t2+t+1)2=[Ostrogradski′sMethod]=−6t(t−1)(t+1)(t2−t+1)(t2+t+1)−6∫t2−5(t2−t+1)(t2+t+1)dt−6∫t2−5(t2−t+1)(t2+t+1)dt==−3∫6t−5t2−t+1dt+3∫6t+5t2+t+1dt==6∫dtt2−t+1−9∫2t−1t2−t+1dt+6∫dtt2+t+1+9∫2t+1t2+t+1dt==43arctan3(2t−1)3−9ln(t2−t+1)+43arctan3(2t+1)3+9ln(t2+t+1)⇒wehave2304∫π/20cosx(cos4x−8cos2x+15)2dx=[−6t(t−1)(t+1)(t2−t+1)(t2+t+1)+43(arctan3(2t−1)3+arctan3(2t+1)3)+9lnt2+t+1t2−t+1]01==2π3+9ln3

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